Let $f:[0,1]→\mathbb R$ and $g:[0,1]→\mathbb R$ be differentiable with $f(0)=g(0)$ and $f'(x)>g'(x)$ for all $x\in [0,1]$. Prove $f(x)>g(x)$ for all $x\in (0,1]$

There only seems to be a related theorem, but only applies if $f'(x)=g'(x)$. I think it might be modifiable to fit this case. The theorem states:

Suppose that $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ and that $f'(x)=g'(x)$ for all $x\in (a,b)$. Then there is a real number $k$ such that $f(x)=g(x)+k$ for all $x\in [a,b]$.

So, we want the case that $f'(x)>g'(x)$. Although, I'm not sure if I would use this theorem for this problem. Do I have to use IVT or construct a new function?