Question

The derivative of f(tan x) w.r.t. g(sec x) at x = π/4 where f`(1) = 2 and g`(√2) =4, is

• (A) $\frac{1}{\sqrt{2}}$
• (B) √2
• (C) 1
• (D) 2

Answer 1

The Correct Option is A

Explanation:
Let $u=f(\tan x)$ and $v=g(\sec x)$$\therefore \frac{du}{dx}=f^{\prime }(\tan x)\cdot {\sec }^{2}x$and $\frac{dv}{dx}=g^{\ast }(\sec x)\cdot \sec x\tan x$$\begin{array}{rl}\therefore \frac{du}{dv}& =\frac{du/dx}{dv/dx}\\ & =\frac{f(\tan x)\cdot {\sec }^{2}x}{8^{\prime }(\sec x)\cdot \sec x\tan x}\\ & =\frac{f^{\prime }(\tan x)\sec x}{g^{\prime }(\sec x)\tan x}\end{array}$At $x=\frac{\pi }{4},\frac{du}{dv}=\frac{f^{\prime }(1){\sec }^{\pi }}{g^{\prime }(\sqrt{2}){\tan }^{\pi }}$$=\frac{2\times \sqrt{2}}{4\times 1}=\frac{1}{\sqrt{2}}$