
The equation of the secant through $(a,f(a))$ and $(b,f(b))$ is \[yf(a)=\frac{f(b)f(a)}{ba}(xa)\] which we can rewrite as \[y=\frac{f(b)f(a)}{ba}(xa)+f(a).\] Let \[g(x)=f(x)\left[\frac{f(b)f(a)}{ba}(xa)+f(a)\right].\] Note that $g(a)=g(b)=0$. Also, $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ since $f$ is. So by Rolle's Theorem there exists $c$ in $(a,b)$ such that $g'(c)=0$. But $\displaystyle g'(x)=f'(x)\frac{f(b)f(a)}{ba}$, so \[g'(c)=f'(c)\frac{f(b)f(a)}{ba}=0.\] Therefore, \[f'(c)=\frac{f(b)f(a)}{ba}\] and the proof is complete. 