# Thread: homogeneous function of degree n

1. ## homogeneous function of degree n

I need to prove that if f is differentiable on R and homogeneous of degree n, then xf'(x)=nf(x).
[Homogeneous of degree n means f(tx)=f(x)t^n for every t>0 such that x and tx are in R.]
Ive tried various rearrangements of the equations but nothing has looked too promising. What should I do?

2. Take $g(t)=f(tx)=f(x)t^n$ then $g$ is obviously differentiable whenever $f$ is, and by the chain rule: $g'(t)=f'(tx)x=nf(x)t^{n-1}$, and taking $t=1$ the result follows.