In: Math

# If f(x) = exg(x), where g(0) = 3 and g'(0) = 4, find f '(0).

If $$f(x)=e^{x} g(x),$$ where $$g(0)=3$$ and $$g(0)=4,$$ find $$f(0)$$

## Solutions

##### Expert Solution

since $$f(x)=e^{x} g(x)$$

Then $$f^{\prime}(x)=e^{x}\left\{\frac{d}{d x}(g(x))\right\}+\left\{\frac{d}{d x}\left(e^{x}\right)\right\} g(x)$$ (by product rule)

$$\begin{array}{c} =e^{x} g^{\prime}(x)+e^{x} g(x) \\ \text { So, } f(0)=e^{0} g^{\prime}(0)+e^{0} g(0)=1 * 4+1 * 3=7 \quad\left(\text { as } e^{0}=1\right) \end{array}$$