Find the derivative of f(sin x) with respect | vedclass

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(A) Let $u = f(\sin x)$ and $v = g(\sec x)$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.Using the chain rule, $\frac{du}{dx} = f'(\sin x) \

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) Let $u = f(\sin x)$ and $v = g(\sec x)$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.Using the chain rule, $\frac{du}{dx} = f'(\sin x) \cdot \cos x$ and $\frac{dv}{dx} = g'(\sec x) \cdot \sec x 	an x$.At $x = \frac{\pi}{4}$, $\sin x = \frac{1}{\sqrt{2}}$, $\cos x = \frac{1}{\sqrt{2}}$, $\sec x = \sqrt{2}$, and $	an x = 1$.Substituting these values: $\frac{du}{dx} = f'(\frac{1}{\sqrt{2}}) \cdot \frac{1}{\sqrt{2}} = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.$\frac{dv}{dx} = g'(\sqrt{2}) \cdot \sqrt{2} \cdot 1 = 1 \cdot \sqrt{2} = \sqrt{2}$.Therefore, $\frac{du}{dv} = \frac{3/\sqrt{2}}{\sqrt{2}} = \frac{3}{2}$.

Find the derivative of $f(\sin x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$, given that $f'(\frac{1}{\sqrt{2}}) = 3$ and $g'(\sqrt{2}) = 1$.

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