Use the mean value theorem of integration to derive the mean value theorem of differentiation I'm trying to use the mean value theorem of integration in order to derive the first mean value theorem. 
I know that the mean value theorem of integration says the following:
Suppose $f:[a,b]\to\mathbb{R}$ is continuous, $g$ is regulated and non-negative and $\int_{a}^b g(y) \, dy>0.$ Then there exists $c\in(a,b)$ such that $$f(c)\int_{a}^b g(y)\, dy=\int_{a}^bf(y)g(y)\, dy$$
My thoughts so far are as follows:
Let $g(y)=1 \quad,\forall y\in(a,b).$
This gives  $$f(c)(b-a)=\int_{a}^bf(y)\, dy$$
Now I'm not sure where to go from here. I know I'm aiming for $$f(b)-f(a)=f'(c)(b-a)$$
 A: It is better to understand both the mean value theorems of integration and differentiation with some focus on the hypotheses under which they hold.
Mean Value Theorem for Derivatives: If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ then there is a point $c \in (a, b)$ such that $$f(b) - f(a) = (b - a)f'(c)$$
Mean Value Theorem for Integrals: If $g$ is continuous on $[a, b]$ then there is a point $c \in [a, b]$ for which $$\int_{a}^{b}g(x)\,dx = (b - a)g(c)$$
Looking at the conclusion of both the theorems it appears that it is possible to prove one theorem using the other by taking $f' = g$. And here we see the fundamental problem. If $f' = g$ then we see that $g$ has been mentioned to be continuous on $[a, b]$ whereas $f'$ need not be continuous on $[a, b]$. In fact it is possible that there is an $f'$ for which $\int_{a}^{b}f'(x)\,dx$ does not even make sense.
Thus it follows that mean value theorem for derivatives is a much stronger result compared to the mean value theorem for integrals. Hence it is not possible to prove the mean value for derivatives using the mean value theorem for integrals.

However if we relax the conditions of mean value theorem for derivatives then it is possible to prove it via mean value theorem for integrals. This relaxed version we state first.
Relaxed Mean Value Theorem for Derivatives: If $f$ is a function such that $f'$ is continuous on $[a, b]$ then there is a point $c \in [a, b]$ for which $$f(b) - f(a) = (b - a)f'(c)$$
Proof via Mean Value Theorem for Integrals: Let $g = f'$ then $g$ is continuous and hence by mean value theorem for integrals we have $$\int_{a}^{b}g(x)\,dx = (b - a)g(c) = (b - a)f'(c)$$ for some point $c \in [a, b]$. By Fundamental Theorem of Calculus we have $$\int_{a}^{b}g(x)\,dx = \int_{a}^{b}f'(x)\,dx = f(b) - f(a)$$ and thus our proof is complete.

Now the surprise!! The above proof is circular and hence constitutes more of a hand-vaving if we think in terms of rigor. The reason is that the Fundamental Theorem of Calculus which links integral to the difference between values of anti-derivative is crucially dependent on and proved via Mean Value Theorem for Derivatives.
It is therefore expected on part of book authors / instructors to avoid such questions in exercises. It is better to mention the similarities between both the theorems and further state explicitly the following:
From the similarities between the two theorems one might expect that they are equivalent in the sense that one can be proved via another, but there are some subtle differences between the two and the mean value theorem for derivatives is much stronger. One can use the mean value theorem for derivatives to prove mean value theorem for integrals but not the other way round.
