Let \(f:\mathbb R\to\mathbb R\) be defined by \(f(x)=xe^{2x}\). Find a general formula for the \(n\)th derivative \(f^{(n)}(x)\), and prove it by induction.
Solution
Denoting \(D^n f \definedas \dd{^n f}{x^n}\) and noting that \(D^1 D^n f = D^{n+1}f\), we have \[\begin{align*} D^0f &= xe^{2x} \\ D^1f &= e^{2x} + 2D^0f \\ D^2f &= 2e^{2x} + 2D^1f \\ D^3f &= 4e^{2x} + 2D^2f \\ \end{align*}\]
and (claim) so we find that \[D^n f = 2^{n-1}e^{2x} + 2D^{n-1}f.\]
This holds for \(n=1\): we have \(D^1 f = e^{2x} + 2xe^{2x} = 2^0e^{2x} + 2D^0 f\), and the inductive step holds exactly because \[ D^{n+1}f = DD^{n}f \\ = D(2^{n-1}e^{2x} + 2D^{n-1}f) \\ =2^n e^{2x} + 2D^n f. \] \(\qed\)
Solution
Goal: Find a closed formula for the \(n\)-th derivative \(f^{(n)}(x)\) of \(f(x) = x e^{2x}\) for all integers \(n \ge 0\), and prove it by induction.
1. Claim: For all \(n \ge 0\), \(f^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x}\).
By mathematical induction on \(n\).
2. Base Case (\(n = 0\)): \(f^{(0)}(x) = f(x) = (2^0 x + 0 \cdot 2^{-1}) e^{2x} = x e^{2x}\).
Direct substitution: \(2^0 = 1\) and \(0 \cdot 2^{-1} = 0\), which matches \(f(x)\).
3. Inductive Step: Assume the statement holds for some integer \(k \ge 0\), i.e., \(f^{(k)}(x) = (2^k x + k 2^{k-1}) e^{2x}\). Then it holds for \(k+1\).Proof:
3.1. Differentiating \(f^{(k)}(x)\) using the product rule:\[f^{(k+1)}(x) = \frac{d}{dx}\left[(2^k x + k 2^{k-1}) e^{2x}\right] = \left(\frac{d}{dx}(2^k x + k 2^{k-1})\right) e^{2x} + (2^k x + k 2^{k-1}) \left(\frac{d}{dx} e^{2x}\right).\]
3.2. Compute the components:\[\frac{d}{dx}(2^k x + k 2^{k-1}) = 2^k, \qquad \frac{d}{dx} e^{2x} = 2 e^{2x}.\]
3.3. Combine and factor out \(e^{2x}\):\[f^{(k+1)}(x) = 2^k e^{2x} + 2(2^k x + k 2^{k-1}) e^{2x} = \left(2 \cdot 2^k x + 2^k + 2 k 2^{k-1}\right) e^{2x}\]\[= \left(2^{k+1} x + 2^k(1 + k)\right) e^{2x} = \left(2^{k+1} x + (k+1) 2^{(k+1)-1}\right) e^{2x}.\]This matches the formula for \(n = k+1\).
4. Alternative derivation using Leibniz’s Rule:Proof:
4.1. Leibniz’s Rule states that \((u v)^{(n)} = \sum_{k=0}^n \binom{n}{k} u^{(n-k)} v^{(k)}\).
4.2. Let \(u(x) = x\) and \(v(x) = e^{2x}\).Then \(u^{(0)}(x) = x\), \(u^{(1)}(x) = 1\), and \(u^{(j)}(x) = 0\) for all \(j \ge 2\).Also \(v^{(k)}(x) = 2^k e^{2x}\) for all \(k \ge 0\).
4.3. The sum has only two non-zero terms (\(k=n\) and \(k=n-1\) for \(u^{(0)}\) and \(u^{(1)}\)):\[f^{(n)}(x) = \binom{n}{0} x (2^n e^{2x}) + \binom{n}{1} (1) (2^{n-1} e^{2x}) = (2^n x + n 2^{n-1}) e^{2x}.\]
5. Conclusion: \(f^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x} = 2^{n-1}(2x + n)e^{2x}\) for all \(n \ge 0\).
4. Q.E.D.