If f(x)=e^{x} g(x),g(0)=4,and g^{prime}(0)=2, | vedclass

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

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$6$

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(B) Given the function $f(x) = e^{x} g(x)$.Applying the product rule for differentiation,we have $f^{\prime}(x) = \frac{d}{dx}(e^{x}) \cdot g(x) + e^{x} \cdot \frac{d}{dx}(g(x))$.This simplifies to $f^{\prime}(x) = e^{x} g(x) + e^{x} g^{\prime}(x) = e^{x} (g(x) + g^{\prime}(x))$.Now,substitute $x = 0$ into the derivative expression:$f^{\prime}(0) = e^{0} (g(0) + g^{\prime}(0))$.Given $g(0) = 4$ and $g^{\prime}(0) = 2$,and knowing $e^{0} = 1$:$f^{\prime}(0) = 1 \cdot (4 + 2) = 6$.

If $f(x)=e^{x} g(x)$,$g(0)=4$,and $g^{\prime}(0)=2$,then $f^{\prime}(0)=$

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