Question

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

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

Answer 1

Answer: (A)

Let $ u=f(\tan x) $ and $ v=g(\sec x) $

$ \therefore $ $ \frac{du}{dx}=f^{'}(\tan x).{{\sec }^{2}}x $ and $ \frac{dv}{dx}=g^{'}(\sec x).\sec x\tan x $

$ \therefore $ $ \frac{du}{dv}=\frac{du/dx}{dv/dx} $

$=\frac{f^{'}(\tan x).{{\sec }^{2}}x}{g^{'}(\sec x).\sec x\tan x} $

$=\frac{f^{'}(\tan x)\sec x}{g^{'}(\sec x)\tan x} $ At $ x=\frac{\pi }{4}, $ $ \frac{du}{dv}=\frac{f^{'}(1)\sec \frac{\pi }{4}}{g^{'}(\sqrt{2})\tan \frac{\pi }{4}} $

$=\frac{2\times \sqrt{2}}{4\times 1}=\frac{1}{\sqrt{2}} $