The derivative of F[f{ phi (x)} ] with respec | vedclass

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(D) To find the derivative of the composite function $y = F[f\{ \phi (x)\} ]$,we apply the chain rule repeatedly.Let $u = f\{ \phi (x)\} $ and $v = \p

Vedclass · Accepted Answer

$F'[f\{ \phi (x)\} ] \cdot f'\{ \phi (x)\} \cdot \phi '(x)$

Vedclass · Answer

(D) To find the derivative of the composite function $y = F[f\{ \phi (x)\} ]$,we apply the chain rule repeatedly.Let $u = f\{ \phi (x)\} $ and $v = \phi (x)$. Then $y = F(u)$ and $u = f(v)$.By the chain rule,$\frac{dy}{dx} = \frac{dF}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.$1$. The derivative of $F$ with respect to $u$ is $F'(u) = F'[f\{ \phi (x)\} ]$.$2$. The derivative of $u = f(v)$ with respect to $v$ is $f'(v) = f'\{ \phi (x)\} $.$3$. The derivative of $v = \phi (x)$ with respect to $x$ is $\phi '(x)$.Multiplying these together,we get $\frac{dy}{dx} = F'[f\{ \phi (x)\} ] \cdot f'\{ \phi (x)\} \cdot \phi '(x)$.

The derivative of $F[f\{ \phi (x)\} ]$ with respect to $x$ is:

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