Let \(f(x) = xe^{2x}\). Writing \(f^{(n)}(x)\) for the \(n\)th derivative of \(f(x)\), prove by induction that \(f^{(n)}(x) = 2^n xe^{2x} + n 2^{n-1}e^{2x}\) for all \(n \geq 0\).
Solution
For \(n=0\), the formula reads \[ f(x)=xe^{2x}=2^0xe^{2x}+0, \] so the base case holds.
Assume \[ f^{(n)}(x)=2^nxe^{2x}+n2^{n-1}e^{2x}. \] Differentiating gives \[ \begin{aligned} f^{(n+1)}(x) &=2^n e^{2x}+2^{n+1}xe^{2x}+n2^n e^{2x}\\ &=2^{n+1}xe^{2x}+(n+1)2^n e^{2x}. \end{aligned} \] This is exactly the claimed formula with \(n\) replaced by \(n+1\). Hence it holds for all \(n\ge0\).