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If $f(x)=e^{x} g(x), g(0)=4, g^{\prime}(0)=2$, then $f^{\prime}(0)=$
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Verified Answer
The correct answer is:
6
Given $f(x)=e^{x} g(x)$
$$
\begin{aligned}
\therefore f^{\prime}(x) &=e^{x} g^{\prime}(x)+g(x) \cdot e^{x} \\
\therefore f^{\prime}(0) &=e^{0} g^{\prime}(0)+g(0) \cdot e^{0} \\
&=2+4=6
\end{aligned}
$$
$$
\begin{aligned}
\therefore f^{\prime}(x) &=e^{x} g^{\prime}(x)+g(x) \cdot e^{x} \\
\therefore f^{\prime}(0) &=e^{0} g^{\prime}(0)+g(0) \cdot e^{0} \\
&=2+4=6
\end{aligned}
$$
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