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

Question

Aicharya · Accepted Answer

Given $f(x)=e^{x} g(x)$ $\begin{aligned} 	herefore f^{\prime}(x) &=e^{x} g^{\prime}(x)+g(x) \cdot e^{x} \ 	herefore f^{\prime}(0) &=e^{0} g^{\prime}(0)+g(0) \cdot e^{0} \ &=2+4=6 \end{aligned}$

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

Solution