Question

If $y={\tan }^{-1}(\sec x-\tan x)$, then $\frac{dy}{dx}$ is equal to:

• A. $2$
• B. $-2$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (D)

$-\frac{1}{2}$

$y={\tan }^{-1}(\sec x-\tan x)$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+(\sec x-\tan x{)}^2}\cdot (\sec x\tan x-{\sec }^{2}x)$, Chain rule

$=\frac{\sec x(\tan x-\sec x)}{1+{\sec }^{2}x-2\sec x\tan x+{\tan }^{2}x}$

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