Question

If $y=\log \tan \sqrt{x}$, then the value of $\frac{dy}{dx}$is:

• A. $\frac{1}{2\sqrt{x}}$
• B. $\frac{{\sec }^2\sqrt{x}}{\left(\sqrt{x}\tan \sqrt{x}\right)}$
• C. $2{\sec }^2\sqrt{x}$
• D. $\frac{{\sec }^2\sqrt{x}}{\left(2\sqrt{x}\tan \sqrt{x}\right)}$

Answer 1

The correct option is D

$\frac{{\sec }^2\sqrt{x}}{\left(2\sqrt{x}\tan \sqrt{x}\right)}$

Find the $\frac{dy}{dx}$:

Given that, $y=\log \tan \sqrt{x}$.

Differentiate $y$ with respect to $x$ using chain rule:

$\begin{array}{rcl}y& =& \log \tan \sqrt{x}\\ \frac{d\mathrm{y}}{d\mathrm{x}}& =& \frac{1}{\tan \sqrt{\mathrm{x}}}\times {\sec }^2\sqrt{\mathrm{x}}\times \frac{1}{2\sqrt{\mathrm{x}}}\left[\frac{d\mathrm{tanx}}{d\mathrm{x}}={\sec }^2\mathrm{x}\right]\\ & =& \frac{{\sec }^2\sqrt{\mathrm{x}}}{2\sqrt{\mathrm{x}}\tan \sqrt{\mathrm{x}}}\end{array}$

Hence, option D is correct.