If f(1)=3 and f^{prime}(1)=2,then the value o | vedclass

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(C) Let $y = \log \left[f\left(e^x+2 x\right)\right]$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{f\left(e^x+2

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$2$

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(C) Let $y = \log \left[f\left(e^x+2 x\right)\right]$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{f\left(e^x+2 x\right)} \cdot f^{\prime}\left(e^x+2 x\right) \cdot \frac{d}{dx}(e^x+2 x)$.$\frac{dy}{dx} = \frac{f^{\prime}\left(e^x+2 x\right) \cdot (e^x+2)}{f\left(e^x+2 x\right)}$.Now,evaluate at $x=0$:At $x=0$,the argument of $f$ is $e^0 + 2(0) = 1 + 0 = 1$.So,$\left. \frac{dy}{dx} \right|_{x=0} = \frac{f^{\prime}(1) \cdot (e^0+2)}{f(1)}$.Given $f(1)=3$ and $f^{\prime}(1)=2$:$\left. \frac{dy}{dx} \right|_{x=0} = \frac{2 \cdot (1+2)}{3} = \frac{2 \cdot 3}{3} = 2$.

If $f(1)=3$ and $f^{\prime}(1)=2$,then the value of $\frac{d}{d x}\left\{\log \left[f\left(e^x+2 x\right)\right]\right\}$ at $x=0$ is

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