1. ## Weird Induction proof

Suppose f:R-->R satisfies f(xy) = xf(y)+yf(x) for all real numbers x,y. Prove that f(1)=0 and that f(u^n)=n*u^(n-1)*f(u) for all natural numbers n and all real numbers u.

I really have no clue where to start on this proof, except that the hint in the problem says that in using induction on n, consider the case u = 0 separately.

Any help would be much appreciated.

2. Originally Posted by morbius27
Suppose f:R-->R satisfies f(xy) = xf(y)+yf(x) for all real numbers x,y. Prove that f(1)=0 and that f(u^n)=n*u^(n-1)*f(u) for all natural numbers n and all real numbers u.

I really have no clue where to start on this proof, except that the hint in the problem says that in using induction on n, consider the case u = 0 separately.

Any help would be much appreciated.
For the first, let $y=1$ then $f(x)=xf(1)+f(x)$ and so $xf(1)=0$ and since $x$ was arbitrary, it follows that $f(1)=0$. For the second suppose you know it for $n$ then $f(u^{n+1})=f(u^nu)=u^nf(u)+uf(u^n)=u^nf(u)+unu^{n-1}f(u)=(n+1)u^nf(u)$