#### Please Solve R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3  Question 20 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{a}{2\left ( 1+a^{2}x^{2} \right )}$

Hint:

\begin{aligned} &\frac{d}{d \mathrm{x}}(\text { constant })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}

Given:

\begin{aligned} &\tan ^{-1}\left\{\frac{\sqrt{1+a^{2} x^{2}}-1}{a x}\right\}, \\ &x \neq 0 \end{aligned}

Solution:

Let,

$y=\tan ^{-1}\left\{\frac{\sqrt{1+a^{2} x^{2}}-1}{a x}\right\}$

$Let\: ax=\tan \theta$

Now

$y=\tan ^{-1}\left \{ \frac{\sqrt{1+\tan ^{2}\theta -1}}{\tan \theta } \right \}$

Using

\begin{aligned} &\sec ^{2} \theta=1+\tan ^{2} \theta \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\sec \theta-1}{\tan \theta}\right\} \\ &\mathrm{y}=\tan ^{-1}\left\{\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right\} \end{aligned}

\begin{aligned} &\text { Using } \tan \theta=\frac{\sin \theta}{\cos \theta}, \sec \theta=\frac{1}{\cos \theta} \\ &y=\tan ^{-1}\left\{\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin d}{\cos \theta}}\right\} \\ &y=\tan ^{-1}\left\{\frac{1-\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right\} \\ &y=\tan ^{-1}\left\{\frac{1-\cos \theta}{\sin \theta}\right\} \end{aligned}

\begin{aligned} &\text { Using } 2 \sin ^{2} \theta=1-\cos 2 \theta \\ &\text { and } 2 \sin \theta \cos \theta=\sin 2 \theta \\ &y=\tan ^{-1}\left\{\frac{2 \sin ^{2} \theta / 2}{2 \sin \theta / 2 \cos \theta / 2}\right\} \\ &y=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\} \end{aligned}

$y=\frac{\theta }{2}$

$y=\frac{1}{2}tan^{-1}ax$

Differentiating with respect to x , We get

\begin{aligned} &\frac{dy}{d x}=\frac{d}{d x}\left(\frac{1}{2} \operatorname{tan}^{-1} x\right)\\ &\text { Using }\\ &\frac{d}{dx}\left(\operatorname{tan}^{-1} x\right) \Rightarrow \frac{1}{1+x^{2}}\\ &\frac{d y}{d x}=\frac{1}{2} \times \frac{1}{1+(a x)^{2}} \times \frac{1}{a}\\ &\frac{d y}{d x}=\frac{a}{2\left(1+a^{2} x^{2}\right)} \end{aligned}