Suppose \(f: \mathbb{R} \to \mathbb{R}\) satisfies \(f(xy) = xf(y) + yf(x)\) for all \(x, y \in \mathbb{R}\). Prove that \(f(1) = 0\) and that \(f(u^n) = n u^{n-1} f(u)\) for all \(n \in \mathbb{N}\) and \(u \in \mathbb{R}\).
Solution
Putting \(x=y=1\) gives \[ f(1)=f(1)+f(1), \] so \(f(1)=0\).
We prove by induction on \(n\ge1\) that \[ f(u^n)=nu^{n-1}f(u). \] For \(n=1\) this is immediate. If it holds for \(n\), then \[ \begin{aligned} f(u^{n+1}) &=f(u^n u)\\ &=u^n f(u)+u f(u^n)\\ &=u^n f(u)+u\bigl(nu^{n-1}f(u)\bigr)\\ &=(n+1)u^n f(u). \end{aligned} \] Thus the formula holds for every positive integer \(n\).