#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 48 maths textbook solution

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

Hint: you must know the rules of solving derivative of inverse trigonometric function

Given: $\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$

Solution:

Let  $y=\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$

Differentiate with respect to x

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)\right\}$

$=\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)^{2}}} \times\left[\frac{\left(x^{2}+a^{2}\right)^{\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(x)-x \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+a^{2}\right)^{\frac{1}{2}}}}{\left[\left(x^{2}+a^{2}\right)^{\frac{1}{2}}\right]^{2}}\right] \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

$=\frac{\sqrt{x^{2}+a^{2}}}{a}\left[\frac{\sqrt{x^{2}+a^{2}}-\frac{x}{2 \sqrt{x^{2}+a^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+a^{2}\right)}{\left(x^{2}+a^{2}\right)}\right]$

$=\frac{\sqrt{x^{2}+a^{2}}}{a\left(x^{2}+a^{2}\right)} \quad\left[\sqrt{x^{2}+a^{2}}-\frac{x}{2 \sqrt{x^{2}+a^{2}}} \times 2 x\right]$

$\Rightarrow \frac{\sqrt{x^{2}+a^{2}}}{a\left(x^{2}+a^{2}\right)} \quad\left[\frac{2\left(x^{2}+a^{2}-x^{2}\right)}{2 \sqrt{x^{2}+a^{2}}}\right]$

\begin{aligned} &\Rightarrow \frac{a^{2}}{a\left(x^{2}+a^{2}\right)} \\\\ &\Rightarrow \frac{a}{\left(x^{2}+a^{2}\right)} \end{aligned}