Question

If $y={\tan }^{-1}(\sec x)$, then $\frac{dy}{dx}$ at $x={\mathrm{cosec}}^{-1}\sqrt{2}$ is equal to

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{2}{3}$
• C. $\frac{\sqrt{2}}{3}$
• D. $\frac{2\sqrt{2}}{3}$

Answer 1

Answer: (C)

$y={\tan }^{-1}(\sec x)$
So, ${\frac{dy}{dx}]}_{x=\frac{\pi }{4}}=\frac{1}{1+{\sec }^{2}x}(\sec x\cdot \tan x)=\frac{\sqrt{2}}{3}$