Take then is obviously differentiable whenever is, and by the chain rule: , and taking the result follows.
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?