Question

The derivative of $ f(\tan x) $ w.r.t. $ g(\sec\, x) $ at $ x=\frac{\pi }{4} $ , where $ f'(1)=2 $ and $ g'(\sqrt{2})=4 $ , is

• A. $ \frac{1}{\sqrt{2}} $
• B. $ \sqrt{2} $
• C. $1$
• D. None of these

Answer 1

Answer: (A)

Let $y=f(\tan x)$ and $u=g(\sec x)$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=f'(\tan x) \sec ^{2} x$
and $\frac{d u}{d x}=g'(\sec x) \cdot \sec x \tan x$
$\therefore \frac{d y}{d u}-\frac{d y / d x}{d u / d x}=\frac{f'(\tan x) \sec ^{2} x}{g'(\sec -x) \sec x \tan x}$
$\therefore \left(\frac{d y}{d x}\right)_{x=\pi / 4}=\frac{f'\left(\tan \frac{\pi}{4}\right)}{g'\left(\sec -\frac{\pi}{4}\right) \sin \frac{-\pi}{4}}$
$=\frac{f'(1) \cdot \sqrt{2}}{g'(\sqrt{2})}=\frac{2 \cdot \sqrt{2}}{4}=\frac{1}{\sqrt{2}}$