proof of extended mean-value theorem

Let $f:[a,b]\to ℝ$ and $g:[a,b]\to ℝ$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Define the function

$$h(x)=f(x)\left(g(b)-g(a)\right)-g(x)\left(f(b)-f(a)\right)-f(a)g(b)+f(b)g(a).$$

Because $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, so is $h$. Furthermore, $h(a)=h(b)=0$, so by Rolle’s theorem there exists a $\xi \in (a,b)$ such that $h^{\prime }(\xi )=0$. This implies that

$$f^{\prime }(\xi )\left(g(b)-g(a)\right)-g^{\prime }(\xi )\left(f(b)-f(a)\right)=0$$

and, if $g(b)\ne g(a)$,

$$\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$