The derivative of \(\mathrm{f}(\sec x)\) with respect to \(\mathrm{g}(\tan x)\) at \(x=\frac{\pi}{4}\), where \(\mathrm{f}^{\prime}(\sqrt{2})=4\) and \(\mathrm{g}^{\prime}(1)=2\), is

\(\begin{aligned} & \text { Let } y=\mathrm{f}(\sec x) \text { and } \mathrm{z}=\mathrm{g}(\tan x) \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{f}^{\prime}(\sec x) \cdot \sec x \tan x \\ & \frac{\mathrm{dz}}{\mathrm{d} x}=\mathrm{g}^{\prime}(\tan x) \cdot \sec ^2 x \\ & \text { Now, } \frac{\mathrm{d} y}{\mathrm{dz}}=\frac{\mathrm{f}^{\prime}(\sec x) \sec x \tan x}{\mathrm{~g}^{\prime}(\tan x) \sec ^2 x} \\ & \frac{\mathrm{d} y}{\mathrm{dz}}=\frac{\mathrm{f}^{\prime}(\sec x) \tan x}{\mathrm{~g}^{\prime}(\tan x) \cdot \sec x} \\ & \left.\frac{d y}{d z}\right|_{x=\frac{\pi}{4}}=\frac{f^{\prime}\left(\sec \frac{\pi}{4}\right) \tan \frac{\pi}{4}}{g^{\prime}\left(\tan \frac{\pi}{4}\right) \sec \frac{\pi}{4}} \\ & =\frac{\mathrm{f}^{\prime}(\sqrt{2}) \cdot(1)}{\mathrm{g}^{\prime}(1) \cdot \sqrt{2}} \Rightarrow \frac{4 \times 1}{2 \sqrt{2}} \\ & =\sqrt{2} \\ & \end{aligned}\)