JEE PYQ: Differentiation Question 16
Question 16 - 2019 (12 Apr Shift 2)
The derivative of $\tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right)$, with respect to $\frac{x}{2}$, where $x \in \left(0, \frac{\pi}{2}\right)$, is:
(1) $1$ (2) $\frac{2}{3}$ (3) $\frac{1}{2}$ (4) $2$
Show Answer
Answer: (4) $2$
Solution
$\tan^{-1}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right) = x - \frac{\pi}{4}$. Derivative w.r.t. $\frac{x}{2}$ is $\frac{1}{1/2} = 2$.
BETA
