Q.10. The derivative of f(tan x) w.r.t g(sec x) at x=\frac{\pi }{4} where f'(1) = 2 and g'(\sqrt{2})=4 is
A. \frac{1}{\sqrt{2}}
B. \sqrt{2}
C. 1
D. 0
Answer :- \frac{1}{\sqrt{2}}
Explanation :-
Let p= f(tan x) and q= g(sec x)
\frac{dp}{dx}=f'(tan \; x)\times sec^{2}x and
\frac{dq}{dx}=g'(sec \; x)\times sec^{2}x\times tan x \therefore \left.\frac{\mathrm{dp}}{\mathrm{dx}}\right|_{x=\frac{\pi}{4}}=\mathrm{f}^{\prime}(1)\times2=4, \left.\frac{\mathrm{dq}}{\mathrm{dx}}\right|_{x=\frac{\pi}{4}}=\mathrm{g}^{\prime}(\sqrt{2})\times\sqrt{2}=4\sqrt{2} \therefore \text{Required Derivative}=\left(\frac{\left.\frac{\mathrm{dp}}{\mathrm{dx}}\right|_{x=\frac{\pi}{4}}}{\left.\frac{\mathrm{dq}}{\mathrm{dx}}\right|_{x=\frac{\pi}{4}}}\right)=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}