#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 10

Answer: $\frac{1}{a x \sqrt{1+a^{2} x^{2}}}$

Hint:   $\text { Let } u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right), v=\sqrt{1+a^{2} x^{2}}$
Given:  $\tan ^{-1}\left(\frac{1+a x}{1-a x}\right) \text { w.r.t } \sqrt{1+a^{2} x^{2}}$

Explanation:  $\text { Let } u=\tan ^{-1}\left(\frac{1+a x}{1-a x}\right)$

\begin{aligned} &v=\sqrt{1+a^{2} x^{2}} \\\\ &\text { Let } a x=\tan \theta \\\\ &u=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \end{aligned}

$=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}\right) \quad\left[\tan \frac{\pi}{4}=1\right]$

\begin{aligned} &=\tan ^{-1} \tan \left(\frac{\pi}{4}+\theta\right) \\\\ &=\frac{\pi}{4}+\theta \end{aligned}

$u=\frac{\pi}{4}+\tan ^{-1}(a x) \; \; \; \; \; \; \; \quad\left[\begin{array}{l} a x=\tan \theta \\ \theta=\tan ^{-1} a x \end{array}\right]$

\begin{aligned} &\frac{d u}{d x}=\frac{1}{1+(a x)^{2}}[a \times 1] \\\\ &=\frac{a}{1+a^{2} x^{2}} \\\\ &v=\sqrt{1+a^{2} x^{2}} \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{1}{2 \sqrt{1+a^{2} x^{2}}} \frac{d}{d x}\left(a^{2} x^{2}\right) \\\\ &=\frac{2 a^{2} x}{2 \sqrt{1+a^{2} x^{2}}}=\frac{a^{2} x}{\sqrt{1+a^{2} x^{2}}} \end{aligned}

$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

$=\frac{\frac{a}{1+a^{2} x^{2}}}{\frac{a^{2} x}{\sqrt{1+a^{2} x^{2}}}}=\frac{1}{a x \sqrt{1+a^{2} x^{2}}}$