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

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(C) Let $y = f(\sec x)$ and $z = g(	an x)$.Using the chain rule,we find the derivatives with respect to $x$:$\frac{dy}{dx} = f^{\prime}(\sec x) \cdot

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$\sqrt{2}$

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(C) Let $y = f(\sec x)$ and $z = g(	an x)$.Using the chain rule,we find the derivatives with respect to $x$:$\frac{dy}{dx} = f^{\prime}(\sec x) \cdot \sec x 	an x$$\frac{dz}{dx} = g^{\prime}(	an x) \cdot \sec^2 x$Now,the derivative of $y$ with respect to $z$ is given by:$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{f^{\prime}(\sec x) \cdot \sec x 	an x}{g^{\prime}(	an x) \cdot \sec^2 x} = \frac{f^{\prime}(\sec x) 	an x}{g^{\prime}(	an x) \sec x}$At $x = \frac{\pi}{4}$,we have $\sec(\frac{\pi}{4}) = \sqrt{2}$ and $	an(\frac{\pi}{4}) = 1$.Substituting these values:$\left. \frac{dy}{dz} ight|_{x=\frac{\pi}{4}} = \frac{f^{\prime}(\sqrt{2}) \cdot 1}{g^{\prime}(1) \cdot \sqrt{2}} = \frac{4 \cdot 1}{2 \cdot \sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

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

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