Consider the function $h\left(x\right)equals;f\left(x\right)\cdot \left(b-a\right)-x\cdot \left(f\left(b\right)-f\left(a\right)\right)$, which is continuous on $\left[a\,b\right]$ because $f\left(x\right)$ is continuous on this interval.

Taking the derivative, we obtain: $hapos;\left(x\right)equals;fapos;\left(x\right)\cdot \left(b-a\right)-\left(f\left(b\right)-f\left(a\right)\right)$ which is differentiable on $\left(a\,b\right)$ because $f\left(x\right)$ is differentiable on this interval.

Note that: $h\left(a\right)equals;f\left(a\right)\cdot \left(b-a\right)-a\cdot \left(f\left(b\right)-f\left(a\right)\right)$

$equals;b\cdot f\left(a\right)-a\cdot f\left(a\right)-a\cdot f\left(b\right)plus;a\cdot f\left(a\right)$

$\=b\cdot f\left(a\right)-a\cdot f\left(b\right)$

$\=b\cdot f\left(b\right)-a\cdot f\left(b\right)-b\cdot f\left(b\right)plus;b\cdot f\left(a\right)$

$\=f\left(b\right)\cdot \left(b-a\right)-b\cdot \left(f\left(b\right)-f\left(a\right)\right)$

$\=h\left(b\right)$

Therefore by Rolle's Theorem, there exists a point $c$ in $\left(a\,b\right)$ such that: $hapos;\left(c\right)equals;0equals;fapos;\left(c\right)\cdot \left(b-a\right)-\left(f\left(b\right)-f\left(a\right)\right)$.

And so, $f\left(b\right)-f\left(a\right)equals;fapos;\left(c\right)\cdot \left(b-a\right)$ which can be rearranged to give us $fapos;\left(c\right)equals;\frac{f\left(b\right)-f\left(a\right)}{b-a}$.