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

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

Aicharya · Accepted Answer

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

The derivative of $\mathrm{f}(\sec x)$ with respect to $\mathrm{g}(\tan x)$ at $x=\frac{\pi}{4}$, where…

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