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

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

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

Vedclass · Answer

(A) Let $p = f(	an x)$ and $q = g(\sec x)$.Using the chain rule,we find the derivatives with respect to $x$:$\frac{dp}{dx} = f^{\prime}(	an x) \cdot \sec^2 x$$\frac{dq}{dx} = g^{\prime}(\sec x) \cdot \sec x 	an x$At $x = \frac{\pi}{4}$,we have $	an x = 1$ and $\sec x = \sqrt{2}$.$\left. \frac{dp}{dx} ight|_{x=\frac{\pi}{4}} = f^{\prime}(1) \cdot (\sqrt{2})^2 = 2 \cdot 2 = 4$.$\left. \frac{dq}{dx} ight|_{x=\frac{\pi}{4}} = g^{\prime}(\sqrt{2}) \cdot (\sqrt{2} \cdot 1) = 4 \cdot \sqrt{2} = 4\sqrt{2}$.The required derivative is $\frac{dp}{dq} = \frac{dp/dx}{dq/dx} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.

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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