Question

The derivative of $f(\tan x)$ w.r.t. $g(\sec x)$ at $x=\frac{\pi }{4}$ , where $f^{\prime }(1)=2$ and $g^{\prime }(\sqrt{2})=4$ , is

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $1$
• D. 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
$\frac{dy}{dx}=f^{\prime }(\tan x){\sec }^{2}x$
and $\frac{du}{dx}=g^{\prime }(\sec x)\cdot \sec x\tan x$
$\therefore \frac{dy}{du}-\frac{dy/dx}{du/dx}=\frac{f^{\prime }(\tan x){\sec }^{2}x}{g^{\prime }(\sec -x)\sec x\tan x}$
$\therefore {\left(\frac{dy}{dx}\right)}_{x=\pi /4}=\frac{f^{\prime }\left(\tan \frac{\pi }{4}\right)}{g^{\prime }\left(\sec -\frac{\pi }{4}\right)\sin \frac{-\pi }{4}}$
$=\frac{f^{\prime }(1)\cdot \sqrt{2}}{g^{\prime }(\sqrt{2})}=\frac{2\cdot \sqrt{2}}{4}=\frac{1}{\sqrt{2}}$