Question

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

• A. $\frac{1}{2\sqrt{x}}$
• B. $\frac{\sec^{2} \sqrt{x}}{\sqrt{x} \tan x}$
• C. $2 \, \sec^2 \sqrt{x}$
• D. $\frac{\sec^{2} \sqrt{x}}{2\sqrt{x} \tan x}$

Answer 1

Answer: (D)

$\frac{\sec^{2} \sqrt{x}}{2\sqrt{x} \tan x}$

Answer 2

Let $y = \log \tan\, \sqrt {x}$ Diff. both side w.r.t 'x' $\frac{dy}{dx}= \frac{1}{\tan \sqrt{x}} . \frac{d}{dx}\left(\tan \sqrt{x}\right)$ $= \frac{1}{\tan \sqrt{x}}.\sec^{2} \sqrt{x}. \frac{1}{2\sqrt{x}}$ $\Rightarrow \frac{dy}{dx} = \frac{1 \sec^{2} \sqrt{x}}{2\sqrt{x} \tan \sqrt{x}}$