Question

If $y=\tan {}^{-1}\left(\sec x+\tan x\right)$, then $\frac{dy}{dx}=$

• A. $1$
• B. $\frac{1}{2}$
• C. $-1$
• D. $0$

Answer 1

The correct option is B $\frac{1}{2}$

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

differentiate both sides wrt $x$

$\frac{dy}{dx}=\frac{d}{dx}\left({\tan }^{-1}\right(\sec x+\tan x\left)\right)$

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

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

$\therefore \frac{dy}{dx}=\frac{1}{2}$