The derivative of $\mathrm{f}( an x)$ 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)

Question

The derivative of $\mathrm{f}(	an x)$ 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)

Aicharya · Accepted Answer

$\begin{array}{ll} & 	ext { Let } \mathrm{p}=\mathrm{f}(	an x) 	ext { and } \mathrm{q}=\mathrm{g}(\sec x) \ 	herefore \quad & \frac{\mathrm{dp}}{\mathrm{d} x}=\mathrm{f}^{\prime}(	an x) 	imes \sec ^2 x 	ext { and } \ & \frac{\mathrm{dq}}{\mathrm{d} x}=\mathrm{g}^{\prime}(\sec x) 	imes \sec x 	an x \ 	herefore \quad & \left.\frac{\mathrm{dp}}{\mathrm{d} x}ight|_{x=\frac{\pi}{4}}=\mathrm{f}^{\prime}(1) 	imes 2=4, \ & \left.\frac{\mathrm{dq}}{\mathrm{d} x}ight|_{x=\frac{\pi}{4}}=\mathrm{g}^{\prime}(\sqrt{2}) 	imes \sqrt{2}=4 \sqrt{2} \ 	herefore \quad & 	ext { Required Derivative }=\left(\left.\frac{\mathrm{dp}}{\left.\mathrm{dx}ight|_{x=\frac{\pi}{4}} ^{\mathrm{dq}}}ight|_{s=\frac{\pi}{4}} ^{\mathrm{d}}ight)=\frac{4}{4 \sqrt{2}}=\frac{1}{\sqrt{2}}\end{array}$

The derivative of $\mathrm{f}(\tan x)$ w.r.t. $\mathrm{g}(\sec x)$ at $x=\frac{\pi}{4}$ where…

Solution