If f(x) = e^x g(x),g(0) = 2,and g'(0) = 1,the | vedclass

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(B) Given the function $f(x) = e^x g(x)$.Applying the product rule for differentiation,we get:$f'(x) = \frac{d}{dx}(e^x) \cdot g(x) + e^x \cdot \frac{

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$3$

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(B) Given the function $f(x) = e^x g(x)$.Applying the product rule for differentiation,we get:$f'(x) = \frac{d}{dx}(e^x) \cdot g(x) + e^x \cdot \frac{d}{dx}(g(x))$$f'(x) = e^x g(x) + e^x g'(x)$Now,substitute $x = 0$ into the derivative expression:$f'(0) = e^0 g(0) + e^0 g'(0)$Since $e^0 = 1$,$g(0) = 2$,and $g'(0) = 1$,we have:$f'(0) = 1 \cdot 2 + 1 \cdot 1$$f'(0) = 2 + 1 = 3$.

If $f(x) = e^x g(x)$,$g(0) = 2$,and $g'(0) = 1$,then $f'(0)$ is

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