The derivative of f(tan x) with respect to g( | vedclass

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Question

(A) Let $y = f(	an x)$ and $z = g(\sec x)$. We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.First,find the derivative of $y$ with respect to $x

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) Let $y = f(	an x)$ and $z = g(\sec x)$. We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.First,find the derivative of $y$ with respect to $x$:$\frac{dy}{dx} = f^{\prime}(	an x) \cdot \sec^2 x$.Next,find the derivative of $z$ with respect to $x$:$\frac{dz}{dx} = g^{\prime}(\sec x) \cdot \sec x 	an x$.Now,calculate the ratio at $x = \frac{\pi}{4}$:$\frac{dy}{dz} = \frac{f^{\prime}(	an x) \cdot \sec^2 x}{g^{\prime}(\sec x) \cdot \sec x 	an x} = \frac{f^{\prime}(	an x) \cdot \sec x}{g^{\prime}(\sec x) \cdot 	an x}$.At $x = \frac{\pi}{4}$,$	an x = 1$ and $\sec x = \sqrt{2}$:$\frac{dy}{dz} = \frac{f^{\prime}(1) \cdot \sqrt{2}}{g^{\prime}(\sqrt{2}) \cdot 1} = \frac{2 \cdot \sqrt{2}}{4 \cdot 1} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

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

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