उत्तर :

$\lim _{x \to 3} x+3=3+3=6$

उत्तर :

$\lim _{x \to \pi}\left(x-\dfrac{22}{7}\right)=\left(\pi-\dfrac{22}{7}\right)$

उत्तर :

$\lim _{r \to 1} \pi r^{2}=\pi\left(1\right)^{2}=\pi$

उत्तर :

$\lim _{x \to 4} \dfrac{4 x+3}{x-2}=\dfrac{4\left(4\right)+3}{4-2}=\dfrac{16+3}{2}=\dfrac{19}{2}$

उत्तर :

$\lim _{x \to-1} \dfrac{x^{10}+x^{5}+1}{x-1}=\dfrac{\left(-1\right)^{10}+\left(-1\right)^{5}+1}{-1-1}=\dfrac{1-1+1}{-2}=-\dfrac{1}{2}$

उत्तर :

$\lim _{x \to 0} \dfrac{\left(x+1\right)^{5}-1}{x}$

$ x+1=y $ रखें ताकि $ y^5 - $ 1 जब $x \to 0 $

इस प्रकार,

$\lim _{x \to 0} \dfrac{\left(x+1\right)^{5}-1}{x}=\lim _{y \to 1} \dfrac{y^{5}-1}{y-1}$

$\hspace{3cm}=\lim _{y \to 1} \dfrac{y^{5}-1^{5}}{y-1}$

$\hspace{3cm}=5 \cdot 1^{5-1}$ $\qquad\left[\because \ \ \lim _{x \to a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]$

$\hspace{3cm}=5$

$\therefore \ \ \lim _{x \to 0} \dfrac{\left(x+5\right)^{5}-1}{x}=5$

उत्तर :

$ x=2 $ पर, दी गई परिमेय फलन का मान $\dfrac{0}{0}$ के रूप में आता है।

$ \begin{aligned}

\therefore \ \ \lim _{x \to 2} \dfrac{3 x^{2}-x-10}{x^{2}-4} & =\lim _{x \to 2} \dfrac{\left(x-2\right)\left(3 x+5\right)}{\left(x-2\right)\left(x+2\right)} \\ \\ & =\lim _{x \to 2} \dfrac{3 x+5}{x+2} \\ \\ & =\dfrac{3\left(2\right)+5}{2+2} \\ \\ & =\dfrac{11}{4} \end{aligned} $

Answer :

$x=2$ के लिए, दिए गए परिमेय फलन का मान $\dfrac{0}{0}$ के रूप में होता है।

$ \begin{aligned} \therefore \ \ \lim _{x \to 3} \dfrac{x^{4}-81}{2 x^{2}-5 x-3} & =\lim _{x \to 3} \dfrac{\left(x-3\right)\left(x+3\right)\left(x^{2}+9\right)}{\left(x-3\right)\left(2 x+1\right)} \\ \\ & =\lim _{x \to 3} \dfrac{\left(x+3\right)\left(x^{2}+9\right)}{2 x+1} \\ \\ & =\dfrac{\left(3+3\right)\left(3^{2}+9\right)}{2\left(3\right)+1} \\ \\ & =\dfrac{6 \times 18}{7} \\ \\ & =\dfrac{108}{7} \end{aligned} $

Answer :

$\lim _{x \to 0} \dfrac{a x+b}{c x+1}=\dfrac{a\left(0\right)+b}{c\left(0\right)+1}=b$

Answer :

$\lim _{z \to 1} \dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

$z=1$ के लिए, दिए गए फलन का मान $\frac{0}{0}$ के रूप में होता है

$z^{\frac{1}{6}}=x$ रखें ताकि $z ^{\frac{1}{6}}- 1 \text{ as } x\to 1 $

इसलिए,

$\lim _{z \to 1} \dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=\lim _{x \to 1} \dfrac{x^{2}-1}{x-1}$

$ \hspace{2.2cm}=\lim _{x \to 1} \dfrac{x^{2}-1^{2}}{x-1} $

$ \hspace{2.2cm}=2.1^{2-1} \quad\left[\because \ \ \lim _{x \to a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right] $

$\hspace{2.2cm}=2$

$\therefore \ \ \lim _{z \to 1} \dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=2$

उत्तर :

$ \begin{aligned} \lim _{x \to 1} \dfrac{a x^{2}+b x+c}{c x^{2}+b x+a} & =\dfrac{a\left(1\right)^{2}+b\left(1\right)+c}{c\left(1\right)^{2}+b\left(1\right)+a} \\ \\ & =\dfrac{a+b+c}{a+b+c} \\ \\ & =1 \quad\left[a+b+c \neq 0\right] \end{aligned} $

उत्तर :

$\lim _{x \to-2} \dfrac{\frac{1}{x}+\frac{1}{2}}{x+2}$

$ x = -2 $ पर, दी गई फलन का मान $\frac{0}{0}$ के रूप में होता है।

अब,

$ \begin{aligned} \lim _{x \to-2} \dfrac{\frac{1}{x}+\frac{1}{2}}{x+2} & =\lim _{x \to-2} \dfrac{\left(\frac{2+x}{2 x}\right)}{x+2} \\ \\ & =\lim _{x \to-2} \frac{1}{2 x} \\ \\ & =\frac{1}{2\left(-2\right)}=\frac{-1}{4} \end{aligned} $

उत्तर :

$\lim _{x \to 0} \dfrac{\sin a x}{b x}$

$ x = 0 $ पर, दी गई फलन का मान $\dfrac{0}{0}$ के रूप में होता है।

अब,

$\lim _{x \to 0} \dfrac{\sin a x}{b x}=\lim _{x \to 0} \dfrac{\sin a x}{a x} \times \dfrac{a x}{b x}$

$\hspace{2.2cm}=\lim _{x \to 0}\left(\dfrac{\sin a x}{a x}\right) \times\left(\dfrac{a}{b}\right)$

$\hspace{2.2cm}=\dfrac{a}{b} \lim _{a x \to 0}\left(\dfrac{\sin a x}{a x}\right) \quad\left[x \to 0 \Rightarrow a x \to 0\right]$

$\hspace{2.2cm}=\dfrac{a}{b} \times 1 \quad\left[\because \ \ \ \lim _{y \to 0} \dfrac{\sin y}{y}=1\right]$

$\hspace{2.2cm}=\dfrac{a}{b}$

उत्तर :

$\lim _{x \to 0} \dfrac{\sin a x}{\sin b x}, a, b \neq 0$

$ x = 0 $ पर, दी गई फलन का मान $\dfrac{0}{0}$ के रूप में होता है।

अब,

$ \begin{aligned} \lim _{x \to 0} \dfrac{\sin a x}{\sin b x} & =\lim _{x \to 0} \dfrac{\left(\dfrac{\sin a x}{a x}\right) \times a x}{\left(\dfrac{\sin b x}{b x}\right) \times b x} \\ \\

& =\left(\dfrac{a}{b}\right) \times \dfrac{\lim _{a x \to 0}\left(\dfrac{\sin a x}{a x}\right)}{\lim _{b x \to 0}\left(\dfrac{\sin b x}{b x}\right)} \\ \\ & x \to 0 \Rightarrow a x \to 0 \ \text{ और } \ x \to 0 \Rightarrow b x \to 0 \\ \\ & =\left(\dfrac{a}{b}\right) \times \dfrac{1}{1} {\left[\because \ \ \ \lim _{y \to 0} \dfrac{\sin y}{y}=1\right]} \\ \\ & =\dfrac{a}{b} \end{aligned} $

Answer :

$\lim _{x \to \pi} \dfrac{\sin \left(\pi-x\right)}{\pi\left(\pi-x\right)}$

यह देखा जाता है कि $x $ $\to$ $\pi \Rightarrow\left(\pi - x\right) \to 0 $

$ \begin{aligned} \therefore \ \ \lim _{x \to \pi} \dfrac{\sin \left(\pi-x\right)}{\pi\left(\pi-x\right)} & =\dfrac{1}{\pi} \lim _{\left(\pi-x\right) \to 0} \dfrac{\sin \left(\pi-x\right)}{\left(\pi-x\right)} \\ \\ & =\dfrac{1}{\pi} \times 1 \quad\left[\because \ \ \ \lim _{y \to 0} \dfrac{\sin y}{y}=1\right] \\ \\ & =\dfrac{1}{\pi} \end{aligned} $

Answer :

$\lim _{x \to 0} \dfrac{\cos x}{\pi-x}=\dfrac{\cos 0}{\pi-0}=\dfrac{1}{\pi}$

Answer :

$\lim _{x \to 0} \dfrac{\cos 2 x-1}{\cos x-1}$

$x=0$ पर, दी गई फलन का मान $\dfrac{0}{0}$ के रूप में होता है

अब,

$ \begin{aligned} \lim _{x \to 0} \dfrac{\cos 2 x-1}{\cos x-1} & =\lim _{x \to 0} \dfrac{1-2 \sin ^{2} x-1}{1-2 \sin ^{2} \frac{x}{2}-1} \quad\left[\because \ \ \ \cos x=1-2 \sin ^{2} \dfrac{x}{2}\right] \\ \\ & =\lim _{x \to 0} \dfrac{\sin ^{2} x}{\sin ^{2} \frac{x}{2}}=\lim _{x \to 0} \dfrac{\left(\frac{\sin ^{2} x}{x^{2}}\right) \times x^{2}}{\left(\dfrac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right) \times \frac{x^{2}}{4}} \\ \\

& =4 \dfrac{\lim _{x \to 0}\left(\frac{\sin ^{2} x}{x^{2}}\right)}{\lim _{x \to 0}\left(\dfrac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right)} \\ \\ & =4 \dfrac{\left(\lim _{x \to 0} \frac{\sin x}{x}\right)^{2}}{\left(\lim _{\frac{x}{2} \to 0} \dfrac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}} \quad\left[x \to 0 \Rightarrow \dfrac{x}{2} \to 0\right] \\ \\ & =4 \left(\dfrac{1^{2}}{1^{2}}\right) \qquad\left[\because \ \ \ \lim _{y \to 0} \frac{\sin y}{y}=1\right] \\ \\ & =4 \end{aligned} $

Answer :

$\lim _{x \to 0} \dfrac{a x+x \cos x}{b \sin x}$

$x=0$ के लिए, दी गई फलन का मान $\dfrac{0}{0}$ के रूप में ले लिया जाता है।

अब,

$ \begin{aligned} \lim _{x \to 0} \dfrac{a x+x \cos x}{b \sin x} & =\dfrac{1}{b} \lim _{x \to 0} \dfrac{x\left(a+\cos x\right)}{\sin x} \\ \\ & =\dfrac{1}{b} \lim _{x \to 0}\left(\dfrac{x}{\sin x}\right) \times \lim _{x \to 0}\left(a+\cos x\right) \\ \\ & =\dfrac{1}{b} \times \dfrac{1}{\left(\lim _{x \to 0} \dfrac{\sin x}{x}\right)} \times \lim _{x \to 0}\left(a+\cos x\right) \\ \\ & =\dfrac{1}{b} \times\left(a+\cos 0\right) \quad\left[\lim _{x \to 0} \dfrac{\sin x}{x}=1\right] \\ \\ & =\dfrac{a+1}{b} \end{aligned} $

Answer :

$\lim _{x \to 0} x \sec x=\lim _{x \to 0} \dfrac{x}{\cos x}=\dfrac{0}{\cos 0}=\dfrac{0}{1}=0$

Answer :

$x=0$ के लिए, दी गई फलन का मान $\dfrac{0}{0}$ के रूप में ले लिया जाता है।

अब,

$ \begin{aligned} \lim _{x \to 0} \dfrac{\sin a x+b x}{a x+\sin b x} & =\lim _{x \to 0} \dfrac{\left(\dfrac{\sin a x}{a x}\right) a x+b x}{a x+b x\left(\dfrac{\sin b x}{b x}\right)} \\ \\ & =\dfrac{\left(\lim _{a x \to 0} \dfrac{\sin a x}{a x}\right) \times \lim _{x \to 0}\left(a x\right)+\lim _{x \to 0} b x}{\lim _{x \to 0} a x+\lim _{x \to 0} b x\left(\lim _{b x \to 0} \dfrac{\sin b x}{b x}\right)} \qquad\left[\because \ \ \text{ As } x \to 0 \Rightarrow a x \to 0 \text{ and } b x \to 0 \right] \\ \\

& =\dfrac{\lim _{x \to 0}\left(a x\right)+\lim _{x \to 0} b x}{\lim _{x \to 0} a x+\lim _{x \to 0} b x} \qquad\left[\because \ \ \lim _{x \to 0} \dfrac{\sin x}{x}=1\right] \\ \\ & =\dfrac{\lim _{x \to 0}\left(a x+b x\right)}{\lim _{x \to 0}\left(a x+b x\right)} \\ \\ & =\lim _{x \to 0}\left(1\right) \\ \\ & =1 \end{aligned} $

Answer :

$x=0$ पर, दी गई फलन का मान $\infty-\infty$ के रूप में होता है।

अब,

$ \begin{aligned} \lim _{x \to 0}\left(\text{cosec} \ x-\cot x\right) & =\lim _{x \to 0}\left(\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}\right) \\ \\ & =\lim _{x \to 0}\left(\dfrac{1-\cos x}{\sin x}\right) \\ \\ & =\lim _{x \to 0} \dfrac{\left(\dfrac{1-\cos x}{x}\right)}{\left(\dfrac{\sin x}{x}\right)} \\ \\ & =\dfrac{\lim _{x \to 0} \dfrac{1-\cos x}{x}}{\lim _{x \to 0} \dfrac{\sin x}{x}} \qquad\left[\because \ \ \lim _{x \to 0} \dfrac{1-\cos x}{x}=0 \text{ and } \lim _{x \to 0} \dfrac{\sin x}{x}=1\right] \\ \\ & =\dfrac{0}{1} =0 \end{aligned} $

Answer :

$\lim _{x \to ({\pi}/{2})} \dfrac{\tan 2 x}{x-\frac{\pi}{2}}$

$x=\dfrac{\pi}{2}$ पर, दी गई फलन का मान $\dfrac{0}{0}$ के रूप में होता है।

अब, $x-\dfrac{\pi}{2}=y \quad x \to \dfrac{\pi}{2}, y \to 0$ को रखें

$\therefore \ \ \lim _{x \to ({\pi}/{2})} \dfrac{\tan 2 x}{x-\frac{\pi}{2}} =\lim _{y \to 0} \dfrac{\tan 2\left(y+\frac{\pi}{2}\right)}{y} $

$\hspace{2.8cm} =\lim _{y \to 0} \dfrac{\tan \left(\pi+2 y\right)}{y} \quad\left[\because \ \ \tan \left(\pi+2 y\right)=\tan 2 y\right] $

$\hspace{2.8cm} =\lim _{y \to 0} \dfrac{\tan 2 y}{y} $

$\hspace{2.8cm} =\lim _{y \to 0} \dfrac{\sin 2 y}{y\cdot\cos 2 y} $

$\hspace{2.8cm} =\lim _{y \to 0}\left(\dfrac{\sin 2 y}{2 y} \times \dfrac{2}{\cos 2 y}\right) $

$\hspace{2.8cm} =\left(\lim _{2 y \to 0} \dfrac{\sin 2 y}{2 y}\right) \times \lim _{y \to 0}\left(\dfrac{2}{\cos 2 y}\right) \quad\left[\because \ \ \lim _{x \to 0} \dfrac{\sin x}{x}=1\right] $

$\hspace{2.8cm} =1 \times \dfrac{2}{\cos 0} $

$\hspace{2.8cm} =1 \times \dfrac{2}{1} $

$\hspace{2.8cm} =2 $

Answer :

दिया गया फलन है

$f\left(x\right)= \begin{cases}2 x+3, & x \leq 0 \\ \\ 3\left(x+1\right), & x>0\end{cases}$

$\lim _{x \to 0^{-}} f\left(x\right)=\lim _{x \to 0}\left(2 x+3\right)=2\left(0\right)+3=3$

$\lim _{x \to 0^{+}} f\left(x\right)=\lim _{x \to 0} 3\left(x+1\right)=3\left(0+1\right)=3$

$\therefore \ \ \lim _{x \to 0^{-}} f\left(x\right)=\lim _{x \to 0^{+}} f\left(x\right)=\lim _{x \to 0} f\left(x\right)=3$

$\lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1} 3\left(x+1\right)=3\left(1+1\right)=6$ $\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1} 3\left(x+1\right)=3\left(1+1\right)=6$

$\therefore \ \ \lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1} f\left(x\right)=6$

Answer :

दिया गया फलन है

$f\left(x\right)=\begin{cases} x^{2}-1, x \leq 1 \\ \\ -x^{2}-1, x>1 \end{cases} $

$\lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1}\left(x^{2}-1\right)=1^{2}-1=1-1=0$

$\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1}\left(-x^{2}-1\right)=-1^{2}-1=-1-1=-2$

यह देखा गया है कि $\lim _{x \to 1^{-}} f\left(x\right) \neq \lim _{x \to 1^{+}} f\left(x\right)$

अतः, $\lim _{x \to 1} f\left(x\right)$ अस्तित्व नहीं रखता।

उत्तर :

दी गई फलन है

$f\left(x\right)= \begin{cases}\dfrac{|x|}{x}, & x \neq 0 \\ \\ 0, & x=0\end{cases}$

$ \begin{matrix} \lim _{x \to 0^{-}} f\left(x\right) & =\lim _{x \to 0^{-}}\left(\dfrac{|x|}{x}\right) & \\ \\ & =\lim _{x \to 0}\left(\dfrac{-x}{x}\right) \quad & \\ \\ & =\lim _{x \to 0}\left(-1\right) \\ \\ & =-1 \\ \\ \lim _{x \to 0^{+}} f\left(x\right) & =\lim _{x \to 0^{+}}\left(\dfrac{|x|}{x}\right) & \\ \\ & =\lim _{x \to 0}\left(\dfrac{x}{x}\right) \\ \\ & =\lim _{x \to 0}\left(1\right) \\ \\ & =1 \end{matrix} $

यह देखा गया है कि $\lim _{x \to 0^{-}} f\left(x\right) \neq \lim _{x \to 0^{+}} f\left(x\right)$।

अतः, $\lim _{x \to 0} f\left(x\right)$ अस्तित्व नहीं रखता।

उत्तर :

दी गई फलन है

$ \begin{aligned} f\left(x\right) & = \begin{cases}\dfrac{x}{|x|}, & x \neq 0 \\ \\ 0, & x=0\end{cases} \\ \\ \lim _{x \to 0^{-}} f\left(x\right) & =\lim _{x \to 0^{-}}\left(\dfrac{x}{|x|}\right) \\ \\ & =\lim _{x \to 0}\left(\dfrac{x}{-x}\right) \quad\left[\text{ जब } x<0,|x|=-x\right] \\ \\ & =\lim _{x \to 0}\left(-1\right) \\ \\ & =-1 \\ \\ \lim _{x \to 0^{+}} f\left(x\right) & =\lim _{x \to 0^{+}}\left(\dfrac{x}{|x|}\right) \\ \\ & =\lim _{x \to 0}\left(\dfrac{x}{x}\right) \quad\left[\text{ जब } x>0,|x|=x\right] \\ \\ & =\lim _{x \to 0}\left(1\right) \\ \\ & =1 \end{aligned} $

यह देखा गया है कि $\lim _{x \to 0^{-}} f\left(x\right) \neq \lim _{x \to 0^{+}} f\left(x\right)$।

अतः, $\lim _{x \to 0} f\left(x\right)$ अस्तित्व नहीं रखता।

उत्तर :

दी गई फलन $f\left(x\right)=|x|-5$ है।

$ \begin{aligned} & \begin{aligned} \lim _{x \to 5^{-}} f\left(x\right) & =\lim _{x \to 5^{-}}\left[|x|-5\right] \\ \\ & =\lim _{x \to 5}\left(x-5\right) \quad \\ \\ & =5-5 \\ \\ & =0 \end{aligned} \\ \\ & \begin{aligned} \lim _{x \to 5^{+}} f\left(x\right) & =\lim _{x \to 5^{+}}\left(|x|-5\right) \\ \\ & =\lim _{x \to 5}\left(x-5\right) \quad \\ \\ & =5-5 \\ \\ & =0 \end{aligned} \\ \\ & \therefore \ \ \lim _{x \to 5^{-}} f\left(x\right)=\lim _{x \to 5^{+}} f\left(x\right)=0 \end{aligned} $

इसलिए, $\lim _{x \to 5} f\left(x\right)=0$

उत्तर :

दी गई फलन

$f\left(x\right)= \begin{cases}a+b x, & x<1 \\ \\ 4, & x=1 \\ \\ b-a x & x>1\end{cases}$

$\lim _{x \to 1^{-}} f\left(x\right)=\lim _{x \to 1}\left(a+b x\right)=a+b$

$\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1}\left(b-a x\right)=b-a$

$f\left(1\right)=4$

दिया गया है कि $\lim _{x \to 1} f\left(x\right)=f\left(1\right)$।

$\therefore \ \ \lim _{x \to 1 ^{-}} f\left(x\right)=\lim _{x \to 1^{+}} f\left(x\right)=\lim _{x \to 1} f\left(x\right)=f\left(1\right)$

$\Rightarrow a+b=4$ और $b-a=4$

इन दोनों समीकरणों को हल करने पर हमें $a=0$ और $b=4$ प्राप्त होते हैं

इसलिए, $a$ और $b$ के संभावित मान क्रमशः 0 और 4 हैं

उत्तर :

दिए गए फलन है $f\left(x\right)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$

$ \begin{aligned} \lim _{x \to a_1} f\left(x\right) & =\lim _{x \to a_1}\left[\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right] \\ \\ & =\left[\lim _{x \to a_1}\left(x-a_1\right)\right]\left[\lim _{x \to a_1}\left(x-a_2\right)\right] \ldots\left[\lim _{x \to a_1}\left(x-a_n\right)\right] \\ \\ & =\left(a_1-a_1\right)\left(a_1-a_2\right) \ldots\left(a_1-a_n\right)=0 \end{aligned} $

$\therefore \ \ \lim _{x \to a_1} f\left(x\right)=0$

अब,

$ \begin{aligned} \lim _{x \to a} f\left(x\right) & =\lim _{x \to a}\left[\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right] \\ \\ & =\left[\lim _{x \to a}\left(x-a_1\right)\right]\left[\lim _{x \to a}\left(x-a_2\right)\right] \ldots\left[\lim _{x \to a}\left(x-a_n\right)\right] \\ \\ & =\left(a-a_1\right)\left(a-a_2\right) \ldots .\left(a-a_n\right) \end{aligned} $

$\therefore \ \ \lim _{x \to a} f\left(x\right)=\left(a-a_1\right)\left(a-a_2\right) \ldots\left(a-a_n\right)$

उत्तर :

दिए गए फलन है $f\left(x\right)= \begin{cases}|x|+1, & x<0 \\ \\ 0, & x=0 \\ \\ |x|-1, & x>0\end{cases}$

जब $a=0$,

$ \begin{aligned} \lim _{x \to 0^{-}} f\left(x\right) & =\lim _{x \to 0^{-}}\left(|x|+1\right) \\ \\ & =\lim _{x \to 0}\left(-x+1\right) \quad\left[\text{ यदि } x<0,|x|=-x\right] \\ \\ & =(0+1) \\ \\ & =1 \end{aligned} $

$\lim _{x \to 0^{+}} f\left(x\right) =\lim _{x \to 0^{+}}\left(|x|-1\right) $

$\hspace{2.1cm} =\lim _{x \to 0}\left(x-1\right) \quad\left[\text{ यदि } x>0,|x|=x\right] $

$\hspace{2.1cm}=0-1 $

$\hspace{2.1cm} =-1 $

यहां देखा जाता है कि $\lim _{x \to 0^{-}} f\left(x\right) \neq \lim _{x \to 0^{+}} f\left(x\right)$

$\therefore \ \ \lim _{x \to 0} f\left(x\right)$ अस्तित्व में नहीं है।

जब $a<0$,

$ \begin{aligned} & \begin{aligned} \lim _{x \to a^{-}} f\left(x\right) & =\lim _{x \to a^{-}}\left(|x|+1\right) \\ \\ & =\lim _{x \to a}\left(-x+1\right) \quad\left[x<a<0 \Rightarrow|x|=-x\right] \\ \\ & =-a+1 \end{aligned} \\ \\ & \begin{aligned} \lim _{x \to a^{+}} f\left(x\right) & =\lim _{x \to a^{+}}\left(|x|+1\right) \\ \\ & =\lim _{x \to a}\left(-x+1\right) \quad\left[a<x<0 \Rightarrow|x|=-x\right] \\ \\ & =-a+1 \end{aligned} \\ \\ & \therefore \ \ \lim _{x \to a^{-}} f\left(x\right)=\lim _{x \to a^{+}} f\left(x\right)=-a+1 \end{aligned} $

इसलिए, $f\left(x\right)$ का सीमा $x=a$ पर अस्तित्व में है, जहां $a<0$ है।

जब $a>0$

$ \begin{aligned} \lim _{x \to a^{-}} f\left(x\right) & =\lim _{x \to a^{-}}\left(|x|-1\right) \\ \\ & =\lim _{x \to a}\left(x-1\right) \quad\left[0<x<a \Rightarrow|x|=x\right] \\ \\ & =a-1 \\ \\ \lim _{x \to a^{+}} f\left(x\right) & =\lim _{x \to a^{+}}\left(|x|-1\right) \\ \\ & =\lim _{x \to a}\left(x-1\right) \quad\left[0<a<x \Rightarrow|x|=x\right] \\ \\ & =a-1 \\ \\ & \therefore \ \ \lim _{x \to a^{-}} f\left(x\right)=\lim _{x \to a^{+}} f\left(x\right)=a-1 \end{aligned} $

इसलिए, $f\left(x\right)$ का सीमा $x=a$ पर अस्तित्व में है, जहां $a>0$ है।

इसलिए, $\lim _{x \to a} f\left(x\right)$ सभी $a \neq 0$ के लिए अस्तित्व में है

Answer :

$\lim _{x \to 1} \dfrac{f\left(x\right)-2}{x^{2}-1}=\pi$

$\Rightarrow \dfrac{\lim _{x \to 1}\left(f\left(x\right)-2\right)}{\lim _{x \to 1}\left(x^{2}-1\right)}=\pi$

$\Rightarrow \lim _{x \to 1}\left(f\left(x\right)-2\right)=\pi \lim _{x \to 1}\left(x^{2}-1\right)$

$\Rightarrow \lim _{x \to 1}\left(f\left(x\right)-2\right)=\pi\left(1^{2}-1\right)$

$\Rightarrow \lim _{x \to 1}\left(f\left(x\right)-2\right)=0$

$\Rightarrow \lim _{x \to 1} f\left(x\right)-\lim _{x \to 1} 2=0$

$\Rightarrow \lim _{x \to 1} f\left(x\right)-2=0$

$\therefore \ \ \lim _{x \to 1} f\left(x\right)=2$

Answer :

दी गई फलन है

$ \begin{aligned} f\left(x \right) & = \begin{cases}m x^{2}+n, & x<0 \\ \\ n x+m, & 0 \leq x \leq 1 \\ \\ n x^{3}+m, & x>1\end{cases} \\ \\ \lim _{x \to 0^{-}} f\left(x \right) & =\lim _{x \to 0}\left(m x^{2}+n \right) \\ \\ & =m\left(0 \right)^{2}+n \\ \\ & =n \\ \\ \lim _{x \to 0^{+}} f\left(x \right) & =\lim _{x \to 0}\left(n x+m \right) \\ \\ & =n\left(0 \right)+m \\ \\ & =m . \end{aligned} $

इसलिए, $\lim _{x \to 0} f\left(x \right)$ के अस्तित्व के लिए $m=n$ होना आवश्यक है।

$ \begin{aligned} \lim _{x \to 1^{-}} f\left(x \right) & =\lim _{x \to 1}\left(n x+m \right) \\ \\ & =n\left(1 \right)+m \\ \\ & =m+n \end{aligned} $

$\lim _{x \to 1^{+}} f\left(x \right)=\lim _{x \to 1}\left(n x^{3}+m \right)$

$ \hspace{2cm}=n\left(1 \right)^{3}+m $

$ \hspace{2cm}=m+n $

$\therefore \ \ \lim _{x \to 1^{-}} f\left(x \right)=\lim _{x \to 1^{+}} f(x)$

$\qquad\qquad\quad f\left(x \right)=\lim _{x \to 1} f\left(x \right)$

इसलिए, $\lim _{x \to 1} f\left(x \right)$ के अस्तित्व के लिए $m$ और $n$ के किनी भी पूर्णांक मानों के लिए यह अस्तित्व में होता है।

उत्तर :

मान लीजिए $f\left(x \right)=x^{2} - 2$. अतः,

$ \begin{aligned} f^{\prime}\left(10 \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(10+h \right)-f\left(10 \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\left[\left(10+h \right)^{2}-2\right]-\left(10^{2}-2 \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{10^{2}+2 \cdot 10 \cdot h+h^{2}-2-10^{2}+2}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{20 h+h^{2}}{h} \\ \\ & =\lim _ {h \rightarrow 0}\left(20+h \right)=\left(20+0 \right)=20 \end{aligned} $

अतः, $x^{2} - 2$ का $x=10$ पर अवकलज $20$ है।

उत्तर :

मान लीजिए, $ \ f\left(x \right)=x$. अतः,

$ \ \ \ \ \begin{aligned} f^{\prime}\left(1 \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(1+h \right)-f\left(1 \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\left(1+h \right)-1}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{h}{h} \\ \\ & =\lim _ {h \rightarrow 0}\left(1 \right) \\ \\ & =1 \end{aligned} $

अतः, $x$ का $x=1$ पर अवकलज $1$ है।

उत्तर :

मान लीजिए $f\left(x \right)=99 x \quad$ अतः,

$ \begin{aligned} f^{\prime}\left(100 \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(100+h \right)-f\left(100 \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{99\left(100+h \right)-99\left(100 \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{99 \times 100+99 h-99 \times 100}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{99 h}{h} \\ \\ & =\lim _ {h \rightarrow 0}\left(99 \right)=99 \end{aligned} $

अतः, $99 x$ का $x=100$ पर अवकलज $99$ है।

उत्तर :

(i) मान लीजिए $f\left(x \right)=x^{3} - 27$

अतः पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\left[\left(x+h \right)^{3}-27\right]-\left(x^{3}-27 \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}-x^{3}}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{h^{3}+3 x^{2} h+3 x h^{2}}{h} \\ \\ & =\lim _ {h \rightarrow 0}\left(h^{2}+3 x^{2}+3 x h \right) \\ \\ & =0+3 x^{2}+0=3 x^{2} \end{aligned} $

(ii) मान लीजिए, $ \ f\left(x \right)=\left(x- 1 \right)\left(x - 2 \right)\quad$

अतः पहले सिद्धांत से,

$ \ f^{\prime}\left(x \right)=\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{\left(x+h-1 \right)\left(x+h-2 \right)-\left(x-1 \right)\left(x-2 \right)}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{\left(x^{2}+h x-2 x+h x+h^{2}-2 h-x-h+2 \right)-\left(x^{2}-2 x-x+2 \right)}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{\left(h x+h x+h^{2}-2 h-h \right)}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0} \dfrac{2 h x+h^{2}-3 h}{h} $

$\qquad\quad =\lim _ {h \rightarrow 0}\left(2 x+h-3 \right) $

$\qquad\quad=\left(2 x+0-3 \right) $

$\qquad\quad =2 x-3 $

(iii) मान लीजिए $f\left(x \right)=\dfrac{1}{x^{2}}$

अतः पहले सिद्धांत से,

$f^{\prime}\left(x \right)=\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h}$

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{\dfrac{1}{\left(x+h \right)^{2}}-\dfrac{1}{x^{2}}}{h} $

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{x^{2}-\left(x+h \right)^{2}}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{x^{2}-x^{2}-h^{2}-2 h x}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-h^{2}-2 h x}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\lim _ {h \rightarrow 0}\left[\dfrac{-h-2 x}{x^{2}\left(x+h \right)^{2}}\right] $

$\hspace{1cm} =\dfrac{0-2 x}{x^{2}\left(x+0 \right)^{2}}=\dfrac{-2}{x^{3}} $

$\hspace{0.2cm} f\left(x \right)=\dfrac{x+1}{x-1}$

$ \text{ (iv) Let } f(x)=\dfrac{x+1}{x-1}\quad$

$ \text{ Accordingly, from the first principle, }$

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\left(\dfrac{x+h+1}{x+h-1}-\dfrac{x+1}{x-1} \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\left(x-1 \right)\left(x+h+1 \right)-\left(x+1 \right)\left(x+h-1 \right)}{\left(x-1 \right)\left(x+h-1 \right)}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\left(x^{2}+h x+x-x-h-1 \right)-\left(x^{2}+h x-x+x+h-1 \right)}{\left(x-1 \right)\left(x+h-1 \right)}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 h}{\left(x-1 \right)\left(x+h-1 \right)}\right] \\ \\ & =\lim _ {h \rightarrow 0}\left[\dfrac{-2}{\left(x-1 \right)\left(x+h-1 \right)}\right] \\ \\ & =\dfrac{-2}{\left(x-1 \right)\left(x-1 \right)}=\dfrac{-2}{\left(x-1 \right)^{2}} \end{aligned} $

Answer :

दिया गया फलन है $f\left(x \right)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\ldots+\dfrac{x^{2}}{2}+x+1$

$\dfrac{d}{d x} f\left(x \right)=\dfrac{d}{d x}\left[\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\ldots+\dfrac{x^{2}}{2}+x+1\right]$

$\dfrac{d}{d x} f\left(x \right)=\dfrac{d}{d x}\left(\dfrac{x^{100}}{100} \right)+\dfrac{d}{d x}\left(\dfrac{x^{99}}{99} \right)+\ldots+\dfrac{d}{d x}\left(\dfrac{x^{2}}{2} \right)+\dfrac{d}{d x}\left(x \right)+\dfrac{d}{d x}\left(1 \right)$

उपयोग करते हुए प्रमेय $\dfrac{d}{d x}\left(x^{n} \right)=n x^{n-1}$, हम प्राप्त करते हैं

$\dfrac{d}{d x} f\left(x \right)=\dfrac{100 x^{99}}{100}+\dfrac{99 x^{98}}{99}+\ldots+\dfrac{2 x}{2}+1+0$

$ \qquad \qquad =x^{99}+x^{98}+\ldots+x+1 $

$\therefore f^{\prime}\left(x \right)=x^{99}+x^{98}+\ldots+x+1$

$ x=0 $ पर,

$ f^{\prime}\left(0 \right)=1 $

$ x=1 $ पर,

$ f^{\prime}\left(1 \right)=1^{99}+1^{98}+\ldots+1+1=\left[1+1+\ldots+1+1\right] _ {\text{पदों के लिए}}=1 \times 100=100 $

इसलिए, $ f^{\prime}\left(1 \right)=100 \times f^{1}\left(0 \right) $

उत्तर :

मान लीजिए $ f\left(x \right)=x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n} $

$\therefore f^{\prime}\left(x \right)=\dfrac{d}{d x}\left(x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n} \right)$

$\hspace{1.3cm}=\dfrac{d}{d x}\left(x^{n} \right)+a \dfrac{d}{d x}\left(x^{n-1} \right)+a^{2} \dfrac{d}{d x}\left(x^{n-2} \right)+\ldots+a^{n-1} \dfrac{d}{d x}\left(x \right)+a^{n} \dfrac{d}{d x}\left(1 \right)$

प्रमेय $\dfrac{d}{d x} x^{n}=n x^{n-1}$ का उपयोग करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} f^{\prime}\left(x \right) & =n x^{n-1}+a\left(n-1 \right) x^{n-2}+a^{2}\left(n-2 \right) x^{n-3}+\ldots+a^{n-1}+a^{n}\left(0 \right) \\ \\ & =n x^{n-1}+a\left(n-1 \right) x^{n-2}+a^{2}\left(n-2 \right) x^{n-3}+\ldots+a^{n-1} \end{aligned} $

उत्तर :

(i) मान लीजिए $ f\left(x \right)=\left(x - a \right) \left(x - b \right) $

$ \begin{aligned} \Rightarrow f\left(x \right) & =x^{2}-\left(a+b \right) x+a b \\ \\ \therefore f^{\prime}\left(x \right) & =\dfrac{d}{d x}\left(x^{2}-\left(a+b \right) x+a b \right) \\ \\ & =\dfrac{d}{d x}\left(x^{2} \right)-\left(a+b \right) \dfrac{d}{d x}\left(x \right)+\dfrac{d}{d x}\left(a b \right) `

\end{aligned} $

उपयोग करते हुए प्रमेय $\dfrac{d}{d x}\left(x^{n} \right)=n x^{n-1}$, हम प्राप्त करते हैं

$f^{\prime}\left(x \right)=2 x-\left(a+b \right)+0=2 x-a-b$

(ii) मान लीजिए $f\left(x \right)=\left(a x^{2}+b \right)^{2}$

$\Rightarrow f\left(x \right)=a^{2} x^{4}+2 a b x^{2}+b^{2}$

$\therefore f^{\prime}\left(x \right)=\dfrac{d}{d x}\left(a^{2} x^{4}+2 a b x^{2}+b^{2} \right)=a^{2} \dfrac{d}{d x}\left(x^{4} \right)+2 a b \dfrac{d}{d x}\left(x^{2} \right)+\dfrac{d}{d x}\left(b^{2} \right)$

उपयोग करते हुए प्रमेय $\dfrac{d}{d x} x^{n}=n x^{n-1}$, हम प्राप्त करते हैं

$ \begin{aligned} f^{\prime}\left(x \right) & =a^{2}\left(4 x^{3} \right)+2 a b\left(2 x \right)+b^{2}\left(0 \right) \\ \\ & =4 a^{2} x^{3}+4 a b x \\ \\ & =4 a x\left(a x^{2}+b \right) \end{aligned} $

(iii)

मान लीजिए $f\left(x \right)=\dfrac{\left(x-a \right)}{\left(x-b \right)}$

$\Rightarrow f^{\prime}\left(x \right)=\dfrac{d}{d x}\left(\dfrac{x-a}{x-b} \right)$

भाग नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x \right) & =\dfrac{\left(x-b \right) \dfrac{d}{d x}\left(x-a \right)-\left(x-a \right) \dfrac{d}{d x}\left(x-b \right)}{\left(x-b \right)^{2}} \\ \\ & =\dfrac{\left(x-b \right)\left(1 \right)-\left(x-a \right)\left(1 \right)}{\left(x-b \right)^{2}} \\ \\ & =\dfrac{x-b-x+a}{\left(x-b \right)^{2}} \\ \\ & =\dfrac{a-b}{\left(x-b \right)^{2}} \end{aligned} $

उत्तर :

मान लीजिए $f\left(x \right)=\dfrac{x^{n}-a^{n}}{x-a}$

$\Rightarrow f^{\prime}\left(x \right)=\dfrac{d}{d x}\left(\dfrac{x^{n}-a^{n}}{x-a} \right)$

भाग नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x \right) & =\dfrac{\left(x-a \right) \dfrac{d}{d x}\left(x^{n}-a^{n} \right)-\left(x^{n}-a^{n} \right) \dfrac{d}{d x}\left(x-a \right)}{\left(x-a \right)^{2}} \\ \\ & =\dfrac{\left(x-a \right)\left(n x^{n-1}-0 \right)-\left(x^{n}-a^{n} \right)}{\left(x-a \right)^{2}} \\ \\ & =\dfrac{n x^{n}-a n x^{n-1}-x^{n}+a^{n}}{\left(x-a \right)^{2}}

\end{aligned} $

Answer :

(i) मान लीजिए $f\left(x \right)=2 x-\dfrac{3}{4}$

$ \begin{aligned} f^{\prime}\left(x \right) & =\dfrac{d}{d x}\left(2 x-\dfrac{3}{4} \right) \\ \\ & =2 \dfrac{d}{d x}\left(x \right)-\dfrac{d}{d x}\left(\dfrac{3}{4} \right) \\ \\ & =2-0 \\ \\ & =2 \end{aligned} $

(ii) मान लीजिए $f\left(x \right)=\left(5 x^{3}+3 x - 1 \right)\left(x- 1 \right)$

Leibnitz गुणन नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x \right) & =\left(5 x^{3}+3 x-1 \right) \dfrac{d}{d x}\left(x-1 \right)+\left(x-1 \right) \dfrac{d}{d x}\left(5 x^{3}+3 x-1 \right) \\ \\ & =\left(5 x^{3}+3 x-1 \right)\left(1 \right)+\left(x-1 \right)\left(5.3 x^{2}+3-0 \right) \\ \\ & =\left(5 x^{3}+3 x-1 \right)+\left(x-1 \right)\left(15 x^{2}+3 \right) \\ \\ & =5 x^{3}+3 x-1+15 x^{3}+3 x-15 x^{2}-3 \\ \\ & =20 x^{3}-15 x^{2}+6 x-4 \end{aligned} $

(iii) मान लीजिए $f \left(x \right)=x^{-3}\left(5+3 x \right)$

Leibnitz गुणन नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x \right) & =x^{-3} \dfrac{d}{d x}\left(5+3 x \right)+\left(5+3 x \right) \dfrac{d}{d x}\left(x^{-3} \right) \\ \\ & =x^{-3}\left(0+3 \right)+\left(5+3 x \right)\left(-3 x^{-3-1} \right) \\ \\ & =x^{-3}\left(3 \right)+\left(5+3 x \right)\left(-3 x^{-4} \right) \\ \\ & =3 x^{-3}-15 x^{-4}-9 x^{-3} \\ \\ & =-6 x^{-3}-15 x^{-4} \\ \\ & =-3 x^{-3}\left(2+\dfrac{5}{x} \right) \\ \\ & =\dfrac{-3 x^{-3}}{x}\left(2 x+5 \right) \\ \\ & =\dfrac{-3}{x^{4}}\left(5+2 x \right) \end{aligned} $

(iv) मान लीजिए $f\left(x \right)=x^{5}\left(3 - 6 x^{- 9} \right)$

Leibnitz गुणन नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x \right) & =x^{5} \dfrac{d}{d x}\left(3-6 x^{-9} \right)+\left(3-6 x^{-9} \right) \dfrac{d}{d x}\left(x^{5} \right) \\ \\

& =x^{5} \lbrace 0-6\left(-9 \right) x^{-9-1} \rbrace +\left(3-6 x^{-9} \right)\left(5 x^{4} \right) \\ \\ & =x^{5}\left(54 x^{-10} \right)+15 x^{4}-30 x^{-5} \\ \\ & =54 x^{-5}+15 x^{4}-30 x^{-5} \\ \\ & =24 x^{-5}+15 x^{4} \\ \\ & =15 x^{4}+\dfrac{24}{x^{5}} \end{aligned} $

(v) अगर $f\left(x \right)=x^{- 4}\left(3 - 4 x ^{-5} \right)$

Leibnitz गुणन नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x \right) & =x^{-4} \dfrac{d}{d x}\left(3-4 x^{-5} \right)+\left(3-4 x^{-5} \right) \dfrac{d}{d x}\left(x^{-4} \right) \\ \\ & =x^{-4} \lbrace 0-4\left(-5 \right) x^{-5-1} \rbrace +\left(3-4 x^{-5} \right)\left(-4 \right) x^{-4-1} \\ \\ & =x^{-4}\left(20 x^{-6} \right)+\left(3-4 x^{-5} \right)\left(-4 x^{-5} \right) \\ \\ & =20 x^{-10}-12 x^{-5}+16 x^{-10} \\ \\ & =36 x^{-10}-12 x^{-5} \\ \\ & =-\dfrac{12}{x^{5}}+\dfrac{36}{x^{10}} \end{aligned} $

(vi) $ \ \ \dfrac{2}{x+1}-\dfrac{x^2}{3x-1} $

$f^{\prime}\left(x \right)=\dfrac{d}{d x}\left(\dfrac{2}{x+1} \right)-\dfrac{d}{d x}\left(\dfrac{x^{2}}{3 x-1} \right)$

भाग नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x \right) & =\left[\dfrac{\left(x+1 \right) \dfrac{d}{d x}\left(2 \right)-2 \dfrac{d}{d x}\left(x+1 \right)}{\left(x+1 \right)^{2}}\right]-\left[\dfrac{\left(3 x-1 \right) \dfrac{d}{d x}\left(x^{2} \right)-x^{2} \dfrac{d}{d x}\left(3 x-1 \right)}{\left(3 x-1 \right)^{2}}\right] \\ \\ & =\left[\dfrac{\left(x+1 \right)\left(0 \right)-2\left(1 \right)}{\left(x+1 \right)^{2}}\right]-\left[\dfrac{\left(3 x-1 \right)\left(2 x \right)-\left(x^{2} \right)\left(3 \right)}{\left(3 x-1 \right)^{2}}\right] \\ \\ & =\dfrac{-2}{\left(x+1 \right)^{2}}-\left[\dfrac{6 x^{2}-2 x-3 x^{2}}{\left(3 x-1 \right)^{2}}\right] \\ \\ & =\dfrac{-2}{\left(x+1 \right)^{2}}-\left[\dfrac{3 x^{2}-2 x}{\left(3 x-1 \right)^{2}}\right] \\ \\ & =\dfrac{-2}{\left(x+1 \right)^{2}}-\dfrac{x\left(3 x-2 \right)}{\left(3 x-1 \right)^{2}} \end{aligned} $

उत्तर :

मान लीजिए $f\left(x \right)=\cos x$.

इसलिए, पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\cos \left(x+h \right)-\cos x}{h} \\ \\ & =\lim _ {h \rightarrow 0}\left[\dfrac{\cos x \cos h-\sin x \sin h-\cos x}{h}\right] \\ \\ & =\lim _ {h \rightarrow 0}\left[\dfrac{-\cos x\left(1-\cos h \right)-\sin x \sin h}{h}\right] \\ \\ & =\lim _ {h \rightarrow 0}\left[\dfrac{-\cos x\left(1-\cos h \right)}{h}-\dfrac{\sin x \sin h}{h}\right] \\ \\ & =-\cos x\left(\lim _ {h \rightarrow 0} \dfrac{1-\cos h}{h} \right)-\sin x \lim _ {h \rightarrow 0}\left(\dfrac{\sin h}{h} \right) \\ \\ & =-\cos x\left(0 \right)-\sin x\left(1 \right) \\ \\ & =-\sin x \\ \\ \therefore \ \ f^{\prime}\left(x \right) & =-\sin x \end{aligned} $

उत्तर :

(i) मान लीजिए $f\left(x \right)=\sin x \cos x$.

इसलिए, पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\sin \left(x+h \right) \cos \left(x+h \right)-\sin x \cos x}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{2 h}\left[2 \sin \left(x+h \right) \cos \left(x+h \right)-2 \sin x \cos x\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{2 h}\left[\sin 2\left(x+h \right)-\sin 2 x\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{2 h}\left[2 \cos \dfrac{2 x+2 h+2 x}{2} \cdot \sin \dfrac{2 x+2 h-2 x}{2}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos \dfrac{4 x+2 h}{2} \sin \dfrac{2 h}{2}\right] \\ \\

& =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(2 x+h \right) \sin h\right] \\ \\ & =\lim _ {h \rightarrow 0} \cos \left(2 x+h \right) \cdot \lim _ {h \rightarrow 0} \dfrac{\sin h}{h} \\ \\ & =\cos \left(2 x+0 \right) \cdot 1 \\ \\ & =\cos 2 x \end{aligned} $

(ii) मान लीजिए $f \left(x \right)=\sec x$.

इस प्रकार, पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{\sec \left(x+h \right)-\sec x}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\cos \left(x+h \right)}-\dfrac{1}{\cos x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos \left(x+h \right)}{\cos x \cos \left(x+h \right)}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2} \right) \sin \left(\dfrac{x-x-h}{2} \right)}{\cos \left(x+h \right)}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\cos \left(x+h \right)}\right] \\ \\ & =\dfrac{1}{\cos x} \lim _ {h \rightarrow 0} \dfrac{\left[\sin \left(\dfrac{2 x+h}{2} \right) \dfrac{\sin \left(\dfrac{h}{2} \right)}{\left(\dfrac{h}{2} \right)}\right]}{\cos \left(x+h \right)} \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {\frac{h}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2} \right)}{\left(\dfrac{h}{2} \right)} \cdot \lim _ {h \rightarrow 0} \dfrac{\sin \left(\dfrac{2 x+h}{2} \right)}{\cos \left(x+h \right)} \\ \\ & =\dfrac{1}{\cos x} \cdot 1 \dfrac{\sin x}{\cos x} \\ \\ & =\sec x \tan x \end{aligned} $

(iii) मान लीजिए $f \left(x \right)=5 \sec x+4 \cos x$.

इस प्रकार, पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{5 \sec \left(x+h \right)+4 \cos \left(x+h \right)-\left[5 \sec x+4 \cos x\right]}{h} \\ \\

& =5 \lim _ {h \rightarrow 0} \dfrac{\left[\sec \left(x+h \right)-\sec x\right]}{h}+4 \lim _ {h \rightarrow 0} \dfrac{\left[\cos \left(x+h \right)-\cos x\right]}{h} \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\cos \left(x+h \right)}-\dfrac{1}{\cos x}\right]+4 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(x+h \right)-\cos x\right] \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos \left(x+h \right)}{\cos x \cos \left(x+h \right)}\right]+4 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos x \cos h-\sin x \sin h-\cos x\right] \\ \\ & =\dfrac{5}{\cos x} \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2} \right) \sin \left(\dfrac{x-x-h}{2} \right)}{\cos \left(x+h \right)}\right]+4 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[-\cos x\left(1-\cos h \right)-\sin x \sin h\right] \\ \\ & =\dfrac{5}{\cos x} \cdot \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\cos \left(x+h \right)}\right]+4\left[-\cos x \lim _ {h \rightarrow 0} \dfrac{\left(1-\cos h \right)}{h}-\sin x \lim _ {h \rightarrow 0} \dfrac{\sin h}{h}\right] \\ \\ & =\dfrac{5}{\cos x} \cdot \lim _ {h \rightarrow 0}\left[\dfrac{\sin \left(\dfrac{2 x+h}{2} \right) \cdot \dfrac{\sin \left(\dfrac{h}{2} \right)}{\dfrac{h}{2}}}{\cos \left(x+h \right)}\right]+4\left[\left(-\cos x \right) \cdot\left(0 \right)-\left(\sin x \right) \cdot 1\right] \\ \\ & =\dfrac{5}{\cos x} \cdot\left[\lim _ {h \rightarrow 0} \dfrac{\sin \left(\dfrac{2 x+h}{2} \right)}{\cos \left(x+h \right)} \cdot \lim _ {h \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2} \right)}{\dfrac{h}{2}}\right]-4 \sin x \\ \\ & =\dfrac{5}{\cos x} \cdot \dfrac{\sin x}{\cos x} \cdot 1-4 \sin x \\ \\ & =5 \sec x \tan x \cdot-4 \sin x \end{aligned} $

(iv) अब $f\left(x \right)=\text{cosec } x$.

इसलिए, पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\

$$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[cosec \ \left(x+h \right)-cosec \ x\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\sin \left(x+h \right)}-\dfrac{1}{\sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin x-\sin \left(x+h \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2} \right) \cdot \sin \left(\dfrac{x-x-h}{2} \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{-\cos \left(\dfrac{2 x+h}{2} \right) \cdot \dfrac{\sin \left(\dfrac{h}{2} \right)}{\left(\dfrac{h}{2} \right)}}{\sin \left(x+h \right) \sin x} \\ \\ & =\lim _ {h \rightarrow 0}\left(\dfrac{-\cos \left(\frac{2 x+h}{2} \right)}{\sin \left(x+h \right) \sin x} \right) \lim _ {\frac{h}{2} \rightarrow 0} \dfrac{\sin \left(\frac{h}{2} \right)}{\left(\frac{h}{2} \right)} \\ \\ & =\left(\dfrac{-\cos x}{\sin x \sin x} \right) \cdot 1 \\ \\ & =-cosec \ x \cot x \end{aligned} $$

(v) मान लीजिए $f\left(x \right)=3 \cot x+5 cosec \ x$.

इसलिए, पहले सिद्धांत से,

$$ \qquad \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{3 \cot \left(x+h \right)+5 cosec \ \left(x+h \right)-3 \cot x-5 cosec \ x}{h} \\ \\ & =3 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cot \left(x+h \right)-\cot x\right]+5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[cosec \ \left(x+h \right)-cosec \ x\right] \qquad …(1) \end{aligned} $$

अब,

$$ \begin{aligned} \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cot \left(x+h \right)-\cot x\right] & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos \left(x+h \right)}{\sin \left(x+h \right)}-\dfrac{\cos x}{\sin x}\right] \\ \\ $$

& =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos \left(x+h \right) \sin x-\cos x \sin \left(x+h \right)}{\sin x \sin \left(x+h \right)}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x-x-h \right)}{\sin x \sin \left(x+h \right)}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(-h \right)}{\sin x \sin \left(x+h \right)}\right] \\ \\ & =-\left(\lim _ {h \rightarrow 0} \dfrac{\sin h}{h} \right) \cdot\left(\lim _ {h \rightarrow 0} \dfrac{1}{\sin x \cdot \sin \left(x+h \right)} \right) \\ \\ & =-1 \cdot \dfrac{1}{\sin x \cdot \sin \left(x+0 \right)}=\dfrac{-1}{\sin ^{2} x}=-cosec \ ^{2} x \quad\quad … (2) \\ \ \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[cosec \ \left(x+h \right)-cosec \ x\right] & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\sin \left(x+h \right)}-\dfrac{1}{\sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin x-\sin \left(x+h \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2} \right) \cdot \sin \left(\dfrac{x-x-h}{2} \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\sin \left(x+h \right) \sin x}\right] \\ \\ & \lim_ {h\rightarrow 0}\left[\dfrac{-\cos \left(\frac{2 x+h}{2} \right) \cdot \dfrac{\sin \left(\frac{h}{2} \right)}{\left(\frac{h}{2} \right)}}{\sin \left(x+h \right) \sin x}\right] \\ \\ & =\lim _ {h \rightarrow 0}\left(\dfrac{-\cos \left(\frac{2 x+h}{2} \right)}{\sin \left(x+h \right) \sin x} \right) \lim _ {\frac{h}{2} \rightarrow 0} \dfrac{\sin \left(\frac{h}{2} \right)}{\left(\frac{h}{2} \right)} \\ \\ & =\left(\dfrac{-\cos x}{\sin x \sin x} \right) \cdot 1 \\ \\ & =-cosec \ x \cot x \qquad\qquad …(3) \end{aligned} $

(1), (2) और (3) से, हम प्राप्त करते हैं

$ f^{\prime}\left(x \right)=-3 \cosec^{2} x-5 \cosec \ x \cot x $

$(vi)$ मान लीजिए, $f\left(x \right)=5 \sin x$ - $6 \cos x+7$.

इसलिए, पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[5 \sin \left(x+h \right)-6 \cos \left(x+h \right)+7-5 \sin x+6 \cos x-7\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[5 \lbrace \sin \left(x+h \right)-\sin x \rbrace -6 \lbrace \cos \left(x+h \right)-\cos x \rbrace \right] \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\sin \left(x+h \right)-\sin x\right]-6 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(x+h \right)-\cos x\right] \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+x}{2} \right) \sin \left(\dfrac{x+h-x}{2} \right)\right]-6 \lim _ {h \rightarrow 0} \dfrac{\cos x \cos h-\sin x \sin h-\cos x}{h} \\ \\ & =5 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h}{2} \right) \sin \dfrac{h}{2}\right]-6 \lim _ {h \rightarrow 0}\left[\dfrac{-\cos x\left(1-\cos h \right)-\sin x \sin h}{h}\right] \\ \\ & =5 \lim _ {h \rightarrow 0}\left(\cos \left(\dfrac{2 x+h}{2} \right) \dfrac{\sin \frac{h}{2}}{\frac{h}{2}} \right)-6 \lim _ {h \rightarrow 0}\left[\dfrac{-\cos x\left(1-\cos h \right)}{h}-\dfrac{\sin x \sin h}{h}\right] \\ \\ & =5\left[\lim _ {h \rightarrow 0} \cos \left(\dfrac{2 x+h}{2} \right)\right]\left[\lim_ {a\rightarrow 0}\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}\right]-6\left[\left(-\cos x \right)\left(\lim _ {h \rightarrow 0} \dfrac{1-\cos h}{h} \right)-\sin x \lim _ {h \rightarrow 0}\left(\dfrac{\sin h}{h} \right)\right] \\ \\ & =5 \cos x \cdot 1-6\left[\left(-\cos x \right) \cdot\left(0 \right)-\sin x \cdot 1\right] \\ \\ & =5 \cos x+6 \sin x \end{aligned} $

(vii) मान लीजिए $f\left(x \right)=2 \tan x$ - $7 \sec x$.

इसलिए, पहले सिद्धांत से,

$ \begin{aligned}

f^{\prime}\left(x \right) & =\lim _ {h \rightarrow 0} \dfrac{f\left(x+h \right)-f\left(x \right)}{h} \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \tan \left(x+h \right)-7 \sec \left(x+h \right)-2 \tan x+7 \sec x\right] \\ \\ & =\lim _ {h \rightarrow 0} \dfrac{1}{h}\left[2 \lbrace \tan \left(x+h \right)-\tan x \rbrace -7 \lbrace \sec \left(x+h \right)-\sec x \rbrace \right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\tan \left(x+h \right)-\tan x\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\sec \left(x+h \right)-\sec x\right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h \right)}{\cos \left(x+h \right)}-\dfrac{\sin x}{\cos x}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\cos \left(x+h \right)}-\dfrac{1}{\cos x}\right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h \right) \cos x-\sin x \cos \left(x+h \right)}{\cos x \cos \left(x+h \right)}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos \left(x+h \right)}{\cos x \cos \left(x+h \right)}\right] \\ \\ & =2 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h-x \right)}{\cos x \cos \left(x+h \right)}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2} \right) \sin \left(\dfrac{x-x-h}{2} \right)}{\cos x \cos \left(x+h \right)}\right] \\ \\ & =2 \lim _ {h \rightarrow 0}\left[\left(\dfrac{\sin h}{h} \right) \dfrac{1}{\cos x \cos \left(x+h \right)}\right]-7 \lim _ {h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2} \right) \sin \left(-\dfrac{h}{2} \right)}{\cos x \cos \left(x+h \right)}\right] \\ \\ & =2\lim _ {h \rightarrow 0} \left(\dfrac{\sin h}{h} \right)\lim _ {h \rightarrow 0} \left(\dfrac{1}{\cos x \cos \left(x+h \right)} \right)-7\lim _ {\frac{h}{2} \rightarrow 0} \left(\dfrac{\sin \frac{h}{2}}{\frac{h}{2}} \right)\lim _ {h \rightarrow 0} \left(\dfrac{\sin \left(\frac{2 x+h}{2} \right)}{\cos x \cos \left(x+h \right)} \right) \\ \\

& =2\times 1 \times \dfrac{1}{\cos x \cos x}-7\times 1\left(\dfrac{\sin x}{\cos x \cos x} \right) \\ \\ & =2 \sec ^{2} x-7 \sec x \tan x \end{aligned} $

उत्तर :

(i) मान लीजिए $f\left(x\right)=- x$. अतः, $f\left(x+h\right)=-\left(x+h\right)$

पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{-\left(x+h\right)-\left(-x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{-x-h+x}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{-h}{h} \\ \\ & =\lim _ {h \to 0}\left(-1\right)=-1 \end{aligned} $

$\mathrm{(ii) \ मान लीजिए,} \quad f \left(x\right)=\left(-x\right)^{-1}=\dfrac{1}{-x}=\dfrac{-1}{x} $

$\text{अतः, } \ f \left(x+h\right)=\dfrac{-1}{\left(x+h\right)} $

पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{-1}{x+h}-\left(\dfrac{-1}{x}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{-1}{x+h}+\dfrac{1}{x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{-x+\left(x+h\right)}{x\left(x+h\right)}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{-x+x+h}{x\left(x+h\right)}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{h}{x\left(x+h\right)}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{x\left(x+h\right)} \\ \\ & =\dfrac{1}{x \cdot x}=\dfrac{1}{x^{2}} \end{aligned} $

$\mathrm{(iii) ~ मान लीजिए,} \quad f\left(x\right)=\sin \left(x+1\right)$

अतः, $f\left(x+h\right)=\sin \left(x+h+1\right)$

पहले सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\

& =\lim _ {h \to 0} \dfrac{1}{h}\left[\sin \left(x+h+1\right)-\sin \left(x+1\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+1+x+1}{2}\right) \sin \left(\dfrac{x+h+1-x-1}{2}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h+2}{2}\right) \sin \left(\dfrac{h}{2}\right)\right] \\ \\ & =\lim _ {h \to 0}\left[\cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right] \\ \\ & =\lim _ {h \to 0} \cos \left(\dfrac{2 x+h+2}{2}\right) \times \lim _ {\frac{h}{2} \to 0} \left[\dfrac{\sin\frac{h}{2}}{\frac{h}{2}}\right] \\ \\ & =\cos \left(\dfrac{2 x+0+2}{2}\right) \times 1 \\ \\ & =\cos \left(x+1\right) \\ \\ f\left(x\right) & =\cos \left(x-\dfrac{\pi}{8}\right) \quad \text{ Accordingly, } \end{aligned} $

प्रथम सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & \ =\lim _ {h \to 0} \dfrac{1}{h}\left[\cos \left(x+h-\dfrac{\pi}{8}\right)-\cos \left(x-\dfrac{\pi}{8}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[-2 \sin \dfrac{\left(x+h-\dfrac{\pi}{8}+x-\dfrac{\pi}{8}\right)}{2} \sin \left(\dfrac{x+h-\dfrac{\pi}{8}-x+\dfrac{\pi}{8}}{2}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[-2 \sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \sin \dfrac{h}{2}\right] \\ \\ & =\lim _ {h \to 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right] \\ \\ & =\lim _ {h \to 0}\left[-\sin \left(\dfrac{2 x+h-\frac{\pi}{4}}{2}\right)\right] \times\lim _ {\frac{h}{2} \to 0}\left(\dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right) \hspace{30px} \left[\because \ \ \text{As } h \to 0 \Rightarrow \dfrac{h}{2} \to 0\right] \end{aligned} $

$ \begin{aligned} & \quad \ =-\sin \left(\dfrac{2 x+0-\dfrac{\pi}{4}}{2}\right) \times 1 \\ \\ & \quad \ =-\sin \left(x-\dfrac{\pi}{8}\right)

\end{aligned} $

उत्तर :

मान लीजिए $f\left(x\right)=x+a$. इस प्रकार, $f\left(x+h\right)=x+h+a$

पहले सिद्धांत के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{x+h+a-x-a}{h} \\ \\ & =\lim _ {h \to 0}\left(\dfrac{h}{h}\right) \\ \\ & =\lim _ {h \to 0}\left(1\right) \\ \\ & =1 \end{aligned} $

उत्तर :

मान लीजिए $f\left(x\right)=\left(p x+q\right)\left(\dfrac{r}{x}+s\right)$

लेब्निज गुणन नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(p x+q\right)\left(\dfrac{r}{x}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)\left(p x+q\right)^{\prime} \\ \\ & =\left(p x+q\right)\left(r x^{-1}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)\left(p\right) \\ \\ & =\left(p x+q\right)\left(-r x^{-2}\right)+\left(\dfrac{r}{x}+s\right) p \\ \\ & =\left(p x+q\right)\left(\dfrac{-r}{x^{2}}\right)+\left(\dfrac{r}{x}+s\right) p \\ \\ & =\dfrac{-p r}{x}-\dfrac{q r}{x^{2}}+\dfrac{p r}{x}+p s \\ \\ & =p s-\dfrac{q r}{x^{2}} \end{aligned} $

उत्तर :

मान लीजिए $f\left(x\right)=\left(a x+b\right)\left(c x+d\right)^{2}$

लेब्निज गुणन नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(a x+b\right) \dfrac{d}{d x}\left(c x+d\right)^{2}+\left(c x+d\right)^{2} \dfrac{d}{d x}\left(a x+b\right) \\ \\ & =\left(a x+b\right) \dfrac{d}{d x}\left(c^{2} x^{2}+2 c d x+d^{2}\right)+\left(c x+d\right)^{2} \dfrac{d}{d x}\left(a x+b\right) \\ \\ & =\left(a x+b\right)\left[\dfrac{d}{d x}\left(c^{2} x^{2}\right)+\dfrac{d}{d x}\left(2 c d x\right)+\dfrac{d}{d x} d^{2}\right]+\left(c x+d\right)^{2}\left[\dfrac{d}{d x} a x+\dfrac{d}{d x} b\right] \\ \\

& =\left(a x+b\right)\left(2 c^{2} x+2 c d\right)+\left(c x+d^{2}\right) a \\ \\ & =2 c\left(a x+b\right)\left(c x+d\right)+a\left(c x+d\right)^{2} \end{aligned} $

Answer :

Let $f\left(x\right)=\dfrac{a x+b}{c x+d}$

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(c x+d\right) \dfrac{d}{d x}\left(a x+b\right)-\left(a x+b\right) \dfrac{d}{d x}\left(c x+d\right)}{\left(c x+d\right)^{2}} \\ \\ & =\dfrac{\left(c x+d\right)\left(a\right)-\left(a x+b\right)\left(c\right)}{\left(c x+d\right)^{2}} \\ \\ & =\dfrac{a c x+a d-a c x-b c}{\left(c x+d\right)^{2}} \\ \\ & =\dfrac{a d-b c}{\left(c x+d\right)^{2}} \end{aligned} $

Answer :

Let $f\left(x\right)=\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=\dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}}=\dfrac{x+1}{x-1}$, where $x \neq 0$

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(x-1\right) \dfrac{d}{d x}\left(x+1\right)-\left(x+1\right) \dfrac{d}{d x}\left(x-1\right)}{\left(x-1\right)^{2}}, x \neq 0,1 \\ \\ & =\dfrac{\left(x-1\right)\left(1\right)-\left(x+1\right)\left(1\right)}{\left(x-1\right)^{2}}, x \neq 0,1 \\ \\ & =\dfrac{x-1-x-1}{\left(x-1\right)^{2}}, x \neq 0,1 \\ \\ & =\dfrac{-2}{\left(x-1\right)^{2}}, x \neq 0,1 \end{aligned} $

Answer :

Let $f\left(x\right)=\dfrac{1}{a x^{2}+b x+c}$

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(a x^{2}+b x+c\right) \dfrac{d}{d x}\left(1\right)-\dfrac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}} \\ \\ & =\dfrac{\left(a x^{2}+b x+c\right)\left(0\right)-\left(2 a x+b\right)}{\left(a x^{2}+b x+c\right)^{2}} \\ \\ & =\dfrac{-\left(2 a x+b\right)}{\left(a x^{2}+b x+c\right)^{2}} \end{aligned}

$

उत्तर :

मान लीजिए $f\left(x\right)=\dfrac{a x+b}{p x^{2}+q x+r}$

भाग नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(p x^{2}+q x+r\right) \dfrac{d}{d x}\left(a x+b\right)-\left(a x+b\right) \dfrac{d}{d x}\left(p x^{2}+q x+r\right)}{\left(p x^{2}+q x+r\right)^{2}} \\ \\ & =\dfrac{\left(p x^{2}+q x+r\right)\left(a\right)-\left(a x+b\right)\left(2 p x+q\right)}{\left(p x^{2}+q x+r\right)^{2}} \\ \\ & =\dfrac{a p x^{2}+a q x+a r-2 a p x^{2}-a q x-2 b p x-b q}{\left(p x^{2}+q x+r\right)^{2}} \\ \\ & =\dfrac{-a p x^{2}-2 b p x+a r-b q}{\left(p x^{2}+q x+r\right)^{2}} \end{aligned} $

उत्तर :

मान लीजिए $f\left(x\right)=\dfrac{p x^{2}+q x+r}{a x+b}$

भाग नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(a x+b\right) \dfrac{d}{d x}\left(p x^{2}+q x+r\right)-\left(p x^{2}+q x+r\right) \dfrac{d}{d x}\left(a x+b\right)}{\left(a x+b\right)^{2}} \\ \\ & =\dfrac{\left(a x+b\right)\left(2 p x+q\right)-\left(p x^{2}+q x+r\right)\left(a\right)}{\left(a x+b\right)^{2}} \\ \\ & =\dfrac{2 a p x^{2}+a q x+2 b p x+b q-a p x^{2}-a q x-a r}{\left(a x+b\right)^{2}} \\ \\ & =\dfrac{a p x^{2}+2 b p x+b q-a r}{\left(a x+b\right)^{2}} \end{aligned} $

उत्तर :

$ \begin{aligned} \text{ मान लीजिए } f\left(x\right) & =\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x \\ \\ f^{\prime}\left(x\right) & =\dfrac{d}{d x}\left(\dfrac{a}{x^{4}}\right)-\dfrac{d}{d x}\left(\dfrac{b}{x^{2}}\right)+\dfrac{d}{d x}\left(\cos x\right) \\ \\ & =a \dfrac{d}{d x}\left(x^{-4}\right)-b \dfrac{d}{d x}\left(x^{-2}\right)+\dfrac{d}{d x}\left(\cos x\right) \\ \\ & =a\left(-4 x^{-5}\right)-b\left(-2 x^{-3}\right)+\left(-\sin x\right) \quad\left[\dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1} \text{ और } \dfrac{d}{d x}\left(\cos x\right)=-\sin x\right] \\ \\

& =\dfrac{-4 a}{x^{5}}+\dfrac{2 b}{x^{3}}-\sin x \end{aligned} $

Answer :

$ \begin{aligned} & \text{ मान लीजिए, } \ f\left(x\right)=4 \sqrt{x}-2 \\ \\ & \begin{aligned} \qquad f^{\prime}\left(x\right) & =\dfrac{d}{d x}\left(4 \sqrt{x}-2\right)=\dfrac{d}{d x}\left(4 \sqrt{x}\right)-\dfrac{d}{d x}\left(2\right) \\ \\ & =4 \dfrac{d}{d x}\left(x^{\frac{1}{2}}\right)-0=4\left(\dfrac{1}{2} x^{\frac{1}{2}-1}\right) \\ \\ & =\left(2 x^{-\frac{1}{2}}\right)=\dfrac{2}{\sqrt{x}} \end{aligned} \end{aligned} $

Answer :

Let $f\left(x\right)=\left(a x+b\right)^{n}$

Accordingly, $f\left(x+h\right)=\left \lbrace a\left(x+h\right)+b\right \rbrace ^{n}=\left(a x+a h+b\right)^{n}$ By first principle,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{\left(a x+a h+b\right)^{n}-\left(a x+b\right)^{n}}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{\left(a x+b\right)^{n}\left(1+\dfrac{a h}{a x+b}\right)^{n}-\left(a x+b\right)^{n}}{h} \\ \\ & =\left(a x+b\right)^{n} \lim _ {h \to 0} \dfrac{\left(1+\dfrac{a h}{a x+b}\right)^{n}-1}{h} \\ \\ & =\left(a x+b\right)^{n} \lim _ {h \to 0} \dfrac{1}{h}\left[\left \lbrace 1+n\left(\dfrac{a h}{a x+b}\right)+\dfrac{n\left(n-1\right)}{\lfloor 2}\left(\dfrac{a h}{a x+b}\right)^{2}+\ldots\right \rbrace -1\right] \\ \\ & =\left(a x+b\right)^{n} \lim _ {h \to 0}\left[\dfrac{na}{ax+b}+\dfrac{n(n-1)}{\lfloor 2}\times \dfrac{a^2h}{(ax+b)^2}\right]+\ldots\left(\text{ Terms containing higher degrees of } h\right) \\ \\ & =\left(a x+b\right)^{n} \lim _ {h \to 0}\left[\dfrac{n a}{\left(a x+b\right)}+\dfrac{n\left(n-1\right) a^{2} h}{\lfloor 2\left(a x+b\right)^{2}}+\ldots\right] \\ \\ & =\left(a x+b\right)^{n}\left[\dfrac{n a}{\left(a x+b\right)}+0\right] \\ \\ & =n a \dfrac{\left(a x+b\right)^{n}}{\left(a x+b\right)}\left[(a x+b)^{n-1}\right]

\end{aligned} $

Answer :

Let $f\left(x\right)=\left(a x+b\right)^{n}\left(c x+d\right)^{m}$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}\left(x\right)=\left(a x+b\right)^{n} \dfrac{d}{d x}\left(c x+d\right)^{m}+\left(c x+d\right)^{m} \dfrac{d}{d x}\left(a x+b\right)^{n} \qquad \ldots{(1)} \end{aligned} $

$\text{Let,} \ \ f_1\left(x\right)=\left(c x+d\right)^{m}$

$\text{Now, } \ \ f_1\left(x+h\right)=\left(c x+c h+d\right)^{m}$

$f_1^{\prime}\left(x\right)=\lim _ {h \to 0} \dfrac{f_1\left(x+h\right)-f_1^{\prime}\left(x\right)}{h}$

$\hspace{1cm}=\lim _ {h \to 0} \dfrac{\left(c x+c h+d\right)^{m}-\left(c x+d\right)^{m}}{h}$

$\hspace{1cm}=\left(c x+d\right)^{m} \lim _ {h \to 0} \dfrac{1}{h}\left[\left(1+\dfrac{c h}{c x+d}\right)^{m}-1\right]$

$\hspace{1cm}=\left(c x+d\right)^{m} \lim _ {h \to 0} \dfrac{1}{h}\left[\left(1+\dfrac{m c h}{\left(c x+d\right)}+\dfrac{m\left(m-1\right)}{2} \dfrac{\left(c^{2} h^{2}\right)}{\left(c x+d\right)^{2}}+\ldots\right)-1\right]$

$\hspace{1cm}=\left(c x+d\right)^{m} \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{m c h}{\left(c x+d\right)}+\dfrac{m\left(m-1\right) c^{2} h^{2}}{2\left(c x+d\right)^{2}}+\ldots\left( \text{ Terms containing higher degrees of} \ h\right)\right]$

$\hspace{1cm}=\left(c x+d\right)^{m} \lim _ {h \to 0}\left[\dfrac{m c}{\left(c x+d\right)}+\dfrac{m\left(m-1\right) c^{2} h}{2\left(c x+d\right)^{2}}+\ldots\right]$

$\hspace{1cm}=\left(c x+d\right)^{m}\left[\dfrac{m c}{c x+d}+0\right]$

$\hspace{1cm}=\dfrac{m c\left(c x+d\right)^{m}}{\left(c x+d\right)}$

$\hspace{1cm}=m c\left(c x+d\right)^{m-1}$

$\dfrac{d}{d x}\left(c x+d\right)^{It}=m c\left(c x+d\right)^{m-1}$

Similarly, $\dfrac{d}{d x}\left(a x+b\right)^{n}=n a\left(a x+b\right)^{n-1}$

Therefore, from $\left(1\right), \left(2\right), and \left(3\right),$ we obtain

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(a x+b\right)^{n}\left \lbrace m c\left(c x+d\right)^{m-1}\right \rbrace +\left(c x+d\right)^{m}\left \lbrace n a\left(a x+b\right)^{n-1}\right \rbrace \\ \\

& =\left(a x+b\right)^{n-1}\left(c x+d\right)^{m-1}\left[m c\left(a x+b\right)+n a\left(c x+d\right)\right] \end{aligned} $

Answer :

मान लीजिए $f\left(x\right)=\sin \left(x+a\right)$

$f\left(x+h\right)=\sin \left(x+h+a\right)$

प्रथम सिद्धांत से,

$ \begin{aligned} f^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{\sin \left(x+h+a\right)-\sin \left(x+a\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right] \\ \\ & =\lim _ {h \to \infty} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right] \\ \\ & =\lim _ {h \to 0}\left[\cos \left(\dfrac{2 x+2 a+h}{2}\right)\left \lbrace \dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right \rbrace \right] \\ \\ & =\lim _ {h \to 0} \cos \left(\dfrac{2 x+2 a+h}{2}\right) \lim _ {\frac{h}{2} \to 0}\left \lbrace \dfrac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right \rbrace \quad\left[\text{ जैसे } h \to 0 \Rightarrow \dfrac{h}{2} \to 0\right] \\ \\ & =\cos \left(\dfrac{2 x+2 a}{2}\right) \times 1 \quad\left[\lim _ {x \to 0} \dfrac{\sin x}{x}=1\right] \\ \\ & =\cos \left(x+a\right) \end{aligned} $

Answer :

$\text{मान लीजिए,}\quad f\left(x\right)=cosec \ x \cot x$

Leibnitz गुणन सूत्र से,

$f^{\prime}\left(x\right)=cosec \ x\left(\cot x\right)^{\prime}+\cot x\left(cosec \ x\right)^{\prime}$

$\text{मान लीजिए,}\quad f_1\left(x\right)=\cot x$. अतः, $f_1\left(x+h\right)=\cot \left(x+h\right)$

प्रथम सिद्धांत से,

$ \begin{aligned} f_1^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f_1\left(x+h\right)-f_1\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{\cot \left(x+h\right)-\cot x}{h} \\ \\

& =\lim _ {h \to 0} \dfrac{1}{h}\left(\dfrac{\cos \left(x+h\right)}{\sin \left(x+h\right)}-\dfrac{\cos x}{\sin x}\right) \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin x \cos \left(x+h\right)-\cos x \sin \left(x+h\right)}{\sin x \sin \left(x+h\right)}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x-x-h\right)}{\sin x \sin \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(-h\right)}{\sin \left(x+h\right)}\right] \\ \\ & =\dfrac{-1}{\sin x} \cdot\left(\lim _ {h \to 0} \dfrac{\sin h}{h}\right)\left(\lim _ {h \to 0} \dfrac{1}{\sin \left(x+h\right)}\right) \\ \\ & =\dfrac{-1}{\sin {x}} \cdot 1 \cdot\left(\dfrac{1}{\sin \left(x+0\right)}\right) \\ \\ & =\dfrac{-1}{\sin {2} x} \\ \\ & =-cosec \ ^{2} x \qquad \ldots{(2)} \end{aligned} $

$\therefore \ \ \left(\cot x\right)^{\prime}=-cosec \ ^{2} x$

अब, बर्बल $f_2\left(x\right)=cosec \ x$। इस प्रकार, $f_2\left(x+h\right)=cosec \ \left(x+h\right)$

पहले सिद्धांत के अनुसार,

$ \begin{aligned} f_2^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{f_2\left(x+h\right)-f_2\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[cosec \ \left(x+h\right)-cosec \ x\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{1}{\sin \left(x+h\right)}-\dfrac{1}{\sin x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin x-\sin \left(x+h\right)}{\sin x \sin \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2}\right) \sin \left(\dfrac{x-x-h}{2}\right)}{\sin \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{.2 \cos \left(\dfrac{2 x+h}{2}\right) \sin \left(\dfrac{-h}{2}\right)}{\sin \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\sin x} \cdot \lim _ {h \to 0}\left[\dfrac{-\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin \left(x+h\right)}\right] \\ \\

& =\dfrac{-1}{\sin x} \cdot \lim _ {h \to 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \lim _ {h \to 0} \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin \left(x+h\right)} \\ \\ & =\dfrac{-1}{\sin x} \cdot 1 \cdot \dfrac{\cos \left(\dfrac{2 x+0}{2}\right)}{\sin \left(x+0\right)} \\ \\ & =\dfrac{-1}{\sin x} \cdot \dfrac{\cos x}{\sin x} \\ \\ & =-\cos ec x \cdot \cot x \qquad \ldots{(3)} \end{aligned} $

$\therefore \ \ \left(cosec \ x\right)^{\prime}=-cosec \ x \cdot \cot x$

समीकरण (1), (2), और (3) से, हम प्राप्त करते हैं

$ \begin{aligned} f^{\prime}\left(x\right) & =cosec \ x\left(-cosec \ ^{2} x\right)+\cot x\left(-cosec \ x \cot x\right) \\ \\ & =-cosec \ ^{3} x-\cot ^{2} x cosec \ x \end{aligned} $

Answer :

मान लीजिए $f\left(x\right)=\dfrac{\cos x}{1+\sin x}$

भाग नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(1+\sin x\right) \dfrac{d}{d x}\left(\cos x\right)-\left(\cos x\right) \dfrac{d}{d x}\left(1+\sin x\right)}{\left(1+\sin x\right)^{2}} \\ \\ & =\dfrac{\left(1+\sin x\right)\left(-\sin x\right)-\left(\cos x\right)\left(\cos x\right)}{\left(1+\sin x\right)^{2}} \\ \\ & =\dfrac{-\sin x-\sin ^{2} x-\cos ^{2} x}{\left(1+\sin x\right)^{2}} \\ \\ & =\dfrac{-\sin x-\left(\sin ^{2} x+\cos ^{2} x\right)}{\left(1+\sin x\right)^{2}} \\ \\ & =\dfrac{-\sin x-1}{\left(1+\sin x\right)^{2}} \\ \\ & =\dfrac{-\left(1+\sin x\right)}{\left(1+\sin x\right)^{2}} \\ \\ & =\dfrac{-1}{\left(1+\sin x\right)} \end{aligned} $

Answer :

मान लीजिए $f\left(x\right)=\dfrac{\sin x+\cos x}{\sin x-\cos x}$

भाग नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(\sin x-\cos x\right) \dfrac{d}{d x}\left(\sin x+\cos x\right)-\left(\sin x+\cos x\right) \dfrac{d}{d x}\left(\sin x-\cos x\right)}{\left(\sin x-\cos x\right)^{2}} \\ \\

& =\dfrac{\left(\sin x-\cos x\right)\left(\cos x-\sin x\right)-\left(\sin x+\cos x\right)\left(\cos x+\sin x\right)}{\left(\sin x-\cos x\right)^{2}} \\ \\ & =\dfrac{-\left(\sin x-\cos x\right)^{2}-\left(\sin x+\cos x\right)^{2}}{\left(\sin x-\cos x\right)^{2}} \\ \\ & =\dfrac{-\left[\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right]}{\left(\sin x-\cos x\right)^{2}} \\ \\ & =\dfrac{-\left[1+1\right]}{\left(\sin x-\cos x\right)^{2}} \\ \\ & =\dfrac{-2}{\left(\sin x-\cos x\right)^{2}} \end{aligned} $

Answer :

$\text{Let,}$

$f\left(x\right)=\dfrac{\sec x-1}{\sec x+1}$

$f\left(x\right)=\dfrac{\dfrac{1}{\cos x}-1}{\dfrac{1}{\cos x}+1}=\dfrac{1-\cos x}{1+\cos x}$

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(1+\cos x\right) \dfrac{d}{d x}\left(1-\cos x\right)-\left(1-\cos x\right) \dfrac{d}{d x}\left(1+\cos x\right)}{\left(1+\cos x\right)^{2}} \\ \\ & =\dfrac{\left(1+\cos x\right)\left(\sin x\right)-\left(1-\cos x\right)\left(-\sin x\right)}{\left(1+\cos x\right)^{2}} \\ \\ & =\dfrac{\sin x+\cos x \sin x+\sin x-\sin x \cos x}{\left(1+\cos x\right)^{2}} \\ \\ & =\dfrac{2 \sin x}{\left(1+\cos x\right)^{2}} \\ \\ & =\dfrac{2 \sin x}{\left(1+\dfrac{1}{\sec x}\right)^{2}}=\dfrac{2 \sin x}{\dfrac{\left(\sec x+1\right)^{2}}{\sec ^{2} x}} \\ \\ & =\dfrac{2 \sin x \sec ^{2} x}{\left(\sec x+1\right)^{2}} \\ \\ & =\dfrac{\dfrac{2 \sin x}{\cos x} \sec x}{\left(\sec x+1\right)^{2}} \\ \\ & =\dfrac{2 \sec x \tan x}{\left(\sec x+1\right)^{2}} \end{aligned} $

Answer :

Let $y=\sin ^{n} x$.

Accordingly, for $n=1, y=\sin x$.

$\therefore \ \ \dfrac{d y}{d x}=\cos x$, i.e., $\dfrac{d}{d x} \sin x=\cos x$

For $n=2, y=\sin ^{2} x$.

$ \begin{aligned} \therefore \ \ \dfrac{d y}{d x} & =\dfrac{d}{d x}\left(\sin x \sin x\right) \\ \\

& =\left(\sin x\right)^{\prime} \sin x+\sin x\left(\sin x\right)^{\prime} \\ \\ & =\cos x \sin x+\sin x \cos x \\ \\ & =2 \sin x \cos x \qquad \ldots{(1)} \end{aligned} $

$ n=3, y=\sin ^{3} x $।

$ \begin{aligned} \therefore \ \ \dfrac{d y}{d x} & =\dfrac{d}{d x}\left(\sin x \sin ^{2} x\right) \\ \\ & =\left(\sin x\right)^{\prime} \sin ^{2} x+\sin x\left(\sin ^{2} x\right)^{\prime} \left[\text{ द्विगुणित उत्पाद नियम द्वारा }\right] \\ \\ & =\cos x \sin ^{2} x+\sin x\left(2 \sin x \cos x\right) \qquad \left[\text{ (1) का उपयोग करते हुए }\right] \\ \\ & =\cos x \sin ^{2} x+2 \sin ^{2} x \cos x & \\ \\ & =3 \sin ^{2} x \cos x & \end{aligned} $

हम दावा करते हैं कि $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{\left(n-1\right)} x \cos x$

मान लीजिए कि हमारा दावा $n=k$ के लिए सत्य है।

अर्थात, $\dfrac{d}{d x}\left(\sin ^{k} x\right)=k \sin ^{\left(k-1\right)} x \cos x$

विचार करें

$ \begin{aligned} \dfrac{d}{d x}\left(\sin ^{k+1} x\right) & =\dfrac{d}{d x}\left(\sin x \sin ^{k} x\right) \\ \\ & =\left(\sin x\right)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k}\right)’ \\ \\ & =\cos x \sin ^{k} x+\sin x\left(k \sin ^{(k-1)}\right) \\ \\ & =\cos x \sin ^{k} x+k \sin ^{k} x \cos x \\ \\ & =\left(k+1\right) \sin ^{k} x \cos x \\ \\ & =\left(\sin x\right)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k} x\right)^{\prime} \qquad \left[\text{ द्विगुणित उत्पाद नियम द्वारा }\right] \\ \\ & =\cos x \sin ^{k} x+\sin x\left(k \sin ^{(k-1)} x \cos x\right) \qquad {\left[\text{ (2) का उपयोग करते हुए }\right]} \end{aligned} $

इस प्रकार, हमारा दावा $n=k+1$ के लिए सत्य है।

अतः, गणितीय आगमन द्वारा, $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{\left(n-1\right)} x \cos x$

उत्तर :

मान लीजिए $f\left(x\right)=\dfrac{a+b \sin x}{c+d \cos x}$

भाग नियम द्वारा,

$
\begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(c+d \cos x\right) \dfrac{d}{d x}\left(a+b \sin x\right)-\left(a+b \sin x\right) \dfrac{d}{d x}\left(c+d \cos x\right)}{\left(c+d \cos x\right)^{2}} \\ \\

& =\dfrac{\left(c+d \cos x\right)\left(b \cos x\right)-\left(a+b \sin x\right)\left(-d \sin x\right)}{\left(c+d \cos x\right)^{2}} \\ \\ & =\dfrac{c b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x}{\left(c+d \cos x\right)^{2}} \\ \\ & =\dfrac{b c \cos x+a d \sin x+b d\left(\cos ^{2} x+\sin ^{2} x\right)}{\left(c+d \cos x\right)^{2}} \\ \\ & =\dfrac{b c \cos x+a d \sin x+b d}{\left(c+d \cos x\right)^{2}} \end{aligned} $

Answer :

Let $f\left(x\right)=\dfrac{\sin \left(x+a\right)}{\cos x}$

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\cos x \dfrac{d}{d x}\left[\sin \left(x+a\right)\right]-\sin \left(x+a\right) \dfrac{d}{d x} \cos x}{\cos ^{2} x} \\ \\ f^{\prime}\left(x\right) & =\dfrac{\cos x \dfrac{d}{d x}\left[\sin \left(x+a\right)\right]-\sin \left(x+a\right)\left(-\sin x\right)}{\cos ^{2} x} \qquad …{(i)} \end{aligned} $

Let $g\left(x\right)=\sin \left(x+a\right)$. Accordingly, $g\left(x+h\right)=\sin \left(x+h+a\right)$

By first principle,

$ \begin{aligned} g^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{g\left(x+h\right)-g\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\sin \left(x+h+a\right)-\sin \left(x+a\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right] \\ \\ & =\lim _ {h \to 0}{ \cos \left( \dfrac { 2 x + 2 a + h } { 2 } \right) \left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]} \quad\left[\text{ As } h \to 0 \Rightarrow \dfrac{h}{2} \to 0\right] \\ \\ & =\lim _ {h \to 0} \cos \left(\dfrac{2 x+2 a+h}{2}\right) \lim _ {\frac{h}{2} \to 0}\left \lbrace \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right \rbrace \quad\left[\lim _ {h \to 0} \dfrac{\sin h}{h}=1\right] \\ \\

& =\left(\cos \dfrac{2 x+2 a}{2}\right) \times 1 \\ \\ & =\cos \left(x+a\right) \qquad …{(ii)} \end{aligned} $

(1) और (2) से, हम प्राप्त करते हैं

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\cos x \cdot \cos \left(x+a\right)+\sin x \sin \left(x+a\right)}{\cos ^{2} x} \\ \\ & =\dfrac{\cos \left(x+a-x\right)}{\cos ^{2} x} \\ \\ & =\dfrac{\cos a}{\cos ^{2} x} \end{aligned} $

Answer :

मान लीजिए $f\left(x\right)=x^{4}\left(5 \sin x-3 \cos x\right)$

उत्पाद नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =x^{4} \dfrac{d}{d x}\left(5 \sin x-3 \cos x\right)+\left(5 \sin x-3 \cos x\right) \dfrac{d}{d x}\left(x^{4}\right) \\ \\ & =x^{4}\left[5 \dfrac{d}{d x}\left(\sin x\right)-3 \dfrac{d}{d x}\left(\cos x\right)\right]+\left(5 \sin x-3 \cos x\right) \dfrac{d}{d x}\left(x^{4}\right) \\ \\ & =x^{4}\left[5 \cos x-3\left(-\sin x\right)\right]+\left(5 \sin x-3 \cos x\right)\left(4 x^{3}\right) \\ \\ & =x^{3}\left[5 x \cos x+3 x \sin x+20 \sin x-12 \cos x\right] \end{aligned} $

Answer:

मान लीजिए $f\left(x\right)=\left(x^{2}+1\right) \cos x$

उत्पाद नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(x^{2}+1\right) \dfrac{d}{d x}\left(\cos x\right)+\cos x \dfrac{d}{d x}\left(x^{2}+1\right) \\ \\ & =\left(x^{2}+1\right)\left(-\sin x\right)+\cos x\left(2 x\right) \\ \\ & =-x^{2} \sin x-\sin x+2 x \cos x \end{aligned} $

Answer :

मान लीजिए $f\left(x\right)=\left(a x^{2}+\sin x\right)\left(p+q \cos x\right)$

उत्पाद नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(a x^{2}+\sin x\right) \dfrac{d}{d x}\left(p+q \cos x\right)+\left(p+q \cos x\right) \dfrac{d}{d x}\left(a x^{2}+\sin x\right) \\ \\ & =\left(a x^{2}+\sin x\right)\left(-q \sin x\right)+\left(p+q \cos x\right)\left(2 a x+\cos x\right) \\ \\

& =-q \sin x\left(a x^{2}+\sin x\right)+\left(p+q \cos x\right)\left(2 a x+\cos x\right) \end{aligned} $

Answer :

मान लीजिए $f\left(x\right)=\left(x+\cos x\right)\left(x-\tan x\right)$

उत्पाद नियम के अनुसार,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(x+\cos x\right) \dfrac{d}{d x}\left(x-\tan x\right)+\left(x-\tan x\right) \dfrac{d}{d x}\left(x+\cos x\right) \\ \\ & =\left(x+\cos x\right)\left[\dfrac{d}{d x}\left(x\right)-\dfrac{d}{d x}\left(\tan x\right)\right]+\left(x-\tan x\right)\left(1-\sin x\right) \\ \\ & =\left(x+\cos x\right)\left[1-\dfrac{d}{d x} \tan x\right]+\left(x-\tan x\right)\left(1-\sin x\right)\qquad …(i) \\ \\ & \text{ मान लीजिए } g\left(x\right)=\tan x \text{। इस प्रकार, } g\left(x+h\right)=\tan \left(x+h\right) \end{aligned} $

प्रथम सिद्धांत के अनुसार,

$ \begin{aligned} g^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{g\left(x+h\right)-g\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0}\left(\dfrac{\tan \left(x+h\right)-\tan x}{h}\right) \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right)}{\cos \left(x+h\right)}-\dfrac{\sin x}{\cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right) \cos x-\sin x \cos \left(x+h\right)}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h-x\right)}{\cos \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin h}{\cos \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\cos {x}} \cdot\left(\lim _ {h \to 0} \dfrac{\sin h}{h}\right) \cdot\left(\lim _ {h \to 0} \dfrac{1}{\cos \left(x+h\right)}\right) \\ \\ & =\dfrac{1}{\cos x} \times 1 \times \dfrac{1}{\cos \left(x+0\right)} \\ \\ & =\dfrac{1}{\cos ^{2} x} \\ \\ & =\sec ^{2} x \qquad … {(ii)} \end{aligned} $

इसलिए, (i) और (ii) से, हम प्राप्त करते हैं

$ \begin{aligned}

f^{\prime}\left(x\right) & =\left(x+\cos x\right)\left(1-\sec ^{2} x\right)+\left(x-\tan x\right)\left(1-\sin x\right) \\ \\ & =\left(x+\cos x\right)\left(-\tan ^{2} x\right)+\left(x-\tan x\right)\left(1-\sin x\right) \\ \\ & =-\tan ^{2} x\left(x+\cos x\right)+\left(x-\tan x\right)\left(1-\sin x\right) \end{aligned} $

Answer :

Let $f\left(x\right)=\dfrac{4 x+5 \sin x}{3 x+7 \cos x}$

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\left(3 x+7 \cos x\right) \dfrac{d}{d x}\left(4 x+5 \sin x\right)-\left(4 x+5 \sin x\right) \dfrac{d}{d x}\left(3 x+7 \cos x\right)}{\left(3 x+7 \cos x\right)^{2}} \\ \\ & =\dfrac{\left(3 x+7 \cos x\right)\left[4 \dfrac{d}{d x}\left(x\right)+5 \dfrac{d}{d x}\left(\sin x\right)\right]-\left(4 x+5 \sin x\right)\left[3 \dfrac{d}{d x} (x)+7 \dfrac{d}{d x} \cos x\right]}{\left(3 x+7 \cos x\right)^{2}} \\ \\ & =\dfrac{\left(3 x+7 \cos x\right)\left(4+5 \cos x\right)-\left(4 x+5 \sin x\right)\left(3-7 \sin x\right)}{\left(3 x+7 \cos x\right)^{2}} \\ \\ & =\dfrac{12 x+15 x \cos x+28 \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35 \sin ^{2} x}{\left(3 x+7 \cos x\right)^{2}} \\ \\ & =\dfrac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{\left(3 x+7 \cos x\right)^{2}} \\ \\ & =\dfrac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{\left(3 x+7 \cos x\right)^{2}} \end{aligned} $

Answer :

Let $f\left(x\right)=\dfrac{x^{2} \cos \left(\dfrac{\pi}{4}\right)}{\sin x}$

By quotient rule,

$ \begin{aligned} f^{\prime}\left(x\right) & =\cos \dfrac{\pi}{4} \cdot\left[\dfrac{\sin x \dfrac{d}{d x}\left(x^{2}\right)-x^{2} \dfrac{d}{d x}\left(\sin x\right)}{\sin ^{2} x}\right] \\ \\ & =\cos \dfrac{\pi}{4} \cdot\left[\dfrac{\sin x \cdot 2 x-x^{2} \cos x}{\sin ^{2} x}\right] \\ \\

& =\dfrac{x \cos \dfrac{\pi}{4}\left[2 \sin x-x \cos x\right]}{\sin ^{2} x} \end{aligned} $

Answer :

मान लीजिए $f\left(x\right)=\dfrac{x}{1+\tan x}$

$ \ f^{\prime}\left(x\right)=\dfrac{\left(1+\tan x\right) \dfrac{d}{d x}\left(x\right)-x \dfrac{d}{d x}\left(1+\tan x\right)}{\left(1+\tan x\right)^{2}}$

$f^{\prime}\left(x\right)=\dfrac{\left(1+\tan x\right)-x \cdot \dfrac{d}{d x}\left(1+\tan x\right)}{\left(1+\tan x\right)^{2}}\qquad …(i)$

मान लीजिए $g\left(x\right)=1+\tan x$. इस प्रकार, $g\left(x+h\right)=1+\tan \left(x+h\right)$.

प्रथम सिद्धांत से,

$ \begin{aligned} g^{\prime}\left(x\right) & =\lim _ {h \to 0} \dfrac{g\left(x+h\right)-g\left(x\right)}{h} \\ \\ & =\lim _ {h \to 0}\left[\dfrac{1+\tan \left(x+h\right)-1-\tan x}{h}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right)}{\cos \left(x+h\right)}-\dfrac{\sin x}{\cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right) \cos x-\sin x \cos \left(x+h\right)}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h-x\right)}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin h}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\left(\lim _ {h \to 0} \dfrac{\sin h}{h}\right) \cdot\left(\lim _ {h \to 0} \dfrac{1}{\cos \left(x+h\right) \cos x}\right) \\ \\ & =1 \times \dfrac{1}{\cos ^{2} x}=\sec ^{2} x \\ \\ \Rightarrow \dfrac{d}{d x} & \left(1+\tan x\right)=\sec ^{2} x \qquad …{(ii)} \end{aligned} $

(1) और (2) से, हम प्राप्त करते हैं

$ f^{\prime}\left(x\right)=\dfrac{1+\tan x-x \sec ^{2} x}{\left(1+\tan x\right)^{2}} $

Answer :

मान लीजिए $f\left(x\right)=\left(x+\sec x\right)\left(x-\tan x\right)$

उत्पाद नियम से,

$ \begin{aligned} f^{\prime}\left(x\right) & =\left(x+\sec x\right) \dfrac{d}{d x}\left(x-\tan x\right)+\left(x-\tan x\right) \dfrac{d}{d x}\left(x+\sec x\right) \\ \\

& =\left(x+\sec x\right)\left[\dfrac{d}{d x}\left(x\right)-\dfrac{d}{d x} \tan x\right]+\left(x-\tan x\right)\left[\dfrac{d}{d x}\left(x\right)+\dfrac{d}{d x} \sec x\right] \\ \\ & =\left(x+\sec x\right)\left[1-\dfrac{d}{d x} \tan x\right]+\left(x-\tan x\right)\left[1+\dfrac{d}{d x} \sec x\right] \qquad …{(i)} \end{aligned} $

मान लीजिए $f_1\left(x\right)=\tan x, f_2\left(x\right)=\sec x$

इस प्रकार, $f_1\left(x+h\right)=\tan \left(x+h\right)$ और $f_2\left(x+h\right)=\sec \left(x+h\right)$

$ \begin{aligned} f_1^{\prime}\left(x\right) & =\lim _ {h \to 0}\left(\dfrac{f_1\left(x+h\right)-f_1\left(x\right)}{h}\right) \\ \\ & =\lim _ {h \to 0}\left(\dfrac{\tan \left(x+h\right)-\tan x}{h}\right) \\ \\ & =\lim _ {h \to 0}\left[\dfrac{\tan \left(x+h\right)-\tan x}{h}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right)}{\cos \left(x+h\right)}-\dfrac{\sin x}{\cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h\right) \cos x-\sin x \cos \left(x+h\right)}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin \left(x+h-x\right)}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\sin h}{\cos \left(x+h\right) \cos x}\right] \\ \\ & =\left(\lim _ {h \to 0} \dfrac{\sin h}{h}\right) \cdot\left(\lim _ {h \to 0} \dfrac{1}{\cos \left(x+h\right) \cos x}\right) \\ \\ & =1 \times \dfrac{1}{\cos ^{2} x}=\sec ^{2} x \\ \\ \Rightarrow \dfrac{d}{d x} & \tan x=\sec ^{2} x \qquad …{(ii)} \end{aligned} $

$ \begin{aligned} f_2^{\prime}\left(x\right) & =\lim _ {h \to 0}\left(\dfrac{f_2\left(x+h\right)-f_2\left(x\right)}{h}\right) \\ \\ & =\lim _ {h \to 0}\left(\dfrac{\sec \left(x+h\right)-\sec x}{h}\right) \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{1}{\cos \left(x+h\right)}-\dfrac{1}{\cos x}\right] \\ \\ & =\lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos \left(x+h\right)}{\cos \left(x+h\right) \cos x}\right] \\ \\

$$ \begin{aligned} & =\dfrac{1}{\cos x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2}\right) \cdot \sin \left(\dfrac{x-x-h}{2}\right)}{\cos \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \to 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2}\right) \cdot \sin \left(\dfrac{-h}{2}\right)}{\cos \left(x+h\right)}\right] \\ \\ & =\dfrac{1}{\cos x} \cdot \lim _ {h \to 0}\left[\dfrac{\sin \left(\dfrac{2 x+h}{2}\right)\left \lbrace \dfrac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right \rbrace }{\cos \left(x+h\right)}\right] \\ \\ & =\sec x \dfrac{\left \lbrace \lim _ {h \to 0} \sin \left(\dfrac{2 x+h}{2}\right)\right \rbrace \left \lbrace \lim _ {\frac{h}{2} \to 0} \dfrac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right \rbrace }{\lim _ {h \to 0} \cos \left(x+h\right)} \\ \\ & =\sec x \cdot \dfrac{\sin x \cdot 1}{\cos x}\qquad …(iii) \\ \\ & \Rightarrow \dfrac{d}{d x} \sec x=\sec x \tan x \end{aligned} $$

समीकरण $(i), \ (ii),$ और $(iii)$ से, हम प्राप्त करते हैं

$f^{\prime}\left(x\right)=\left(x+\sec x\right)\left(1-\sec ^{2} x\right)+\left(x-\tan x\right)\left(1+\sec x \tan x\right)$

Answer :

मान लीजिए $f\left(x\right)=\dfrac{x}{\sin ^{n} x}$

भाग विधि के अनुसार,

$f^{\prime}\left(x\right)=\dfrac{\sin ^{n} x \dfrac{d}{d x} (x)-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

इसको आसानी से दिखाया जा सकता है कि $\dfrac{d}{d x} \sin ^{n} x=n \sin ^{n-1} x \cos x$

इसलिए,

$ \begin{aligned} f^{\prime}\left(x\right) & =\dfrac{\sin ^{n} x \dfrac{d}{d x} (x)-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x} \\ \\ & =\dfrac{\sin ^{n} x \cdot 1-x\left(n \sin ^{n-1} x \cos x\right)}{\sin ^{2 n} x} \\ \\ & =\dfrac{\sin ^{n-1} x\left(\sin x-n x \cos x\right)}{\sin ^{2 n} x} \\ \\ & =\dfrac{\sin x-n x \cos x}{\sin ^{n+1} x} \end{aligned} $