If f(1) = 3 and f'(1) = 2,then find the value | vedclass

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(C) Let $y = \log f(e^x + 2x)$.Applying the chain rule,the derivative is:$\frac{dy}{dx} = \frac{1}{f(e^x + 2x)} \cdot f'(e^x + 2x) \cdot \frac{d}{dx}(

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$2$

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(C) Let $y = \log f(e^x + 2x)$.Applying the chain rule,the derivative is:$\frac{dy}{dx} = \frac{1}{f(e^x + 2x)} \cdot f'(e^x + 2x) \cdot \frac{d}{dx}(e^x + 2x)$$\frac{dy}{dx} = \frac{f'(e^x + 2x) \cdot (e^x + 2)}{f(e^x + 2x)}$Now,evaluate at $x = 0$:At $x = 0$,the argument of the function is $e^0 + 2(0) = 1 + 0 = 1$.Substituting $x = 0$ into the derivative expression:$\left. \frac{dy}{dx} \right|_{x=0} = \frac{f'(1) \cdot (e^0 + 2)}{f(1)}$Given $f(1) = 3$ and $f'(1) = 2$:$\left. \frac{dy}{dx} \right|_{x=0} = \frac{2 \cdot (1 + 2)}{3} = \frac{2 \cdot 3}{3} = 2$.

If $f(1) = 3$ and $f'(1) = 2$,then find the value of $\frac{d}{dx} \{ \log f(e^x + 2x) \}$ at $x = 0$.

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