Question

The derivative of $f(tanx)$ w.r.t. $g(secx)$ at $x=\pi /4$ where $f^{\prime }(1)=2$ and $g^{\prime }(\sqrt{2})=4$ 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 $v=g(\sec x)$

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

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

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

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

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