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

\(\text { Let } \mathrm{p}=\mathrm{f}(\tan x) \text { and } \mathrm{q}=\mathrm{g}(\sec x) \)
\( \therefore \frac{\mathrm{dp}}{\mathrm{d} x}=\mathrm{f}^{\prime}(\tan x) \times \sec ^2 x \text { and } \)
\( \frac{\mathrm{dq}}{\mathrm{d} x}=\mathrm{g}^{\prime}(\sec x) \times \sec x \tan x \)
\( \therefore\left.\frac{\mathrm{dp}}{\mathrm{d} x}\right|_{x=\frac{\pi}{4}}=\mathrm{f}^{\prime}(1) \times 2=4, \)
\( \left.\frac{\mathrm{dq}}{\mathrm{d} x}\right|_{x=\frac{\pi}{4}}=\mathrm{g}^{\prime}(\sqrt{2}) \times \sqrt{2}=4 \sqrt{2} \)
\( \therefore \text {Required Derivative }=\left(\left.\frac{\mathrm{dp}}{\left.\mathrm{dx}\right|_{x=\frac{\pi}{4}} ^{\mathrm{dq}}}\right|_{s=\frac{\pi}{4}} ^{\mathrm{d}}\right)=\frac{4}{4 \sqrt{2}}=\frac{1}{\sqrt{2}}\)