## Calculus 8th Edition

$$f^{(n)}(x)=(x+n)e^{x}$$
Given $$f(x)= xe^{x},\ \ f^{(n)}(x)=(x+n)e^{x}$$ At $n=1$ $$f'(x)= xe^x+e^x=(x+1)e^{x}$$ Let at $n=k$, $$f^{(k)}(x)=(x+k)e^{x}$$ Differentiate with respect to $x$ \begin{align*} f^{(k+1)}(x)&=(x+k)e^{x}+e^{x}\\ &=(x+k+1)e^{x} \end{align*} Thus is true for $k=n+1$ and $$f^{(n)}(x)=(x+n)e^{x}$$