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If $f(x)=e^{x} g(x), g(0)=2, g^{\prime}(0)=1$, then $f^{\prime}(0)$
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Verified Answer
The correct answer is:
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$f^{\prime}(x)=e^{x} g^{\prime}(x)+e^{x} g(x)$
$$
\begin{aligned}
\Rightarrow \quad f^{\prime}(0)=e^{0} \cdot g^{\prime}(0)+e^{0} g(0) \\
=1 \cdot 1+1 \cdot 2 \\
\left[\because\left\{g^{\prime}(0)=1 \text { and } g(0)=2\right\}\right] \\
=1+2=3
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow \quad f^{\prime}(0)=e^{0} \cdot g^{\prime}(0)+e^{0} g(0) \\
=1 \cdot 1+1 \cdot 2 \\
\left[\because\left\{g^{\prime}(0)=1 \text { and } g(0)=2\right\}\right] \\
=1+2=3
\end{aligned}
$$
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