#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 76

Answer: $\frac{2}{\pi }$

Hint: you must know the rules of solving derivative of trigonometric functions

Given: $f(x)=\sqrt{\tan \sqrt{x}}$

Find: $f^{'}\left(\frac{\pi^{2}}{16}\right)$

Solution:

$f(x)=\sqrt{\tan (\sqrt{x})}$

Differentiate with respect to x,

$f^{\prime}(x)=\frac{1}{2 \sqrt{\tan (\sqrt{x})}} \frac{d}{d x} \tan \sqrt{x}$

$=\frac{1}{2 \sqrt{\tan (\sqrt{x})}} \sec ^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}}$

$f^{\prime}(x) \Rightarrow \frac{\sec ^{2} \sqrt{x}}{4 \sqrt{x \tan (\sqrt{x})}}$

Now, $f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{\sec ^{2} \sqrt{\frac{\pi^{2}}{16}}}{4 \sqrt{\frac{\pi^{2}}{16} \tan \left(\sqrt{\frac{\pi^{2}}{16}}\right)}}$

$=\frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{4\left(\frac{\pi}{4}\right) \sqrt{\tan \left(\frac{\pi}{4}\right)}}$

$\Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi \times(1)} \quad\left[B u t \tan \frac{\pi}{4} \quad \Rightarrow 1\right]$

\begin{aligned} &\Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi} \\\\ &\Rightarrow \frac{\left(\sec \left(\frac{\pi}{4}\right)\right)^{2}}{\pi} \end{aligned}

\begin{aligned} &=\frac{(\sqrt{2})^{2}}{\pi} \\\\ &f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{2}{\pi} \end{aligned}