Question

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

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

Now, $\frac{df\left(tan\,x\right)}{dg\left(sec\,x\right)}$
$=\frac{f '\left(tan\,x\right)sec^{2}\,x}{g'\left(sec\,x\right)sec\,x\,tan\,x}$
$=\frac{f '\left(tan\,x\right)sec\,x}{g'\left(sec\,x\right)tan\,x}$
$\frac{df \left(tan\,x\right)}{dg\left(sec\,x\right)}\bigg|_{at \,x=\pi/4}$ $=\frac{f '\left(1\right)\sqrt{2}}{g'\left(\sqrt{2}\right)\cdot1}$
$=\frac{2\sqrt{2}}{4\cdot1}=\frac{1}{\sqrt{2}}$