Prove there exists $c,d\in(a,b)$ such that $\frac{f'(c)}{f'(d)}=\frac{e^b-e^a}{b-a}e^{-d}$ Assume that $f$ is differentiable on $(a,b)$ and continuous on $[a,b]$, $f'(x)\neq0$.
Obviously it's related to the mean value theorem, so I was thinking
Let $g(x)=e^xf(x)$ and applying the mean value theorem, but that didn't work out well--does the $f'(x)\neq0$ mean that strict monotonicity will be needed to prove this? Or is it just because of the denominator in the question?
Any halp would be appreciated! Thanks！
 A: By the MVT, there exists some $c \in (a, b)$ such that
$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$
Also, consider $g(x) = f(\ln(x))$, on the interval $[e^a, e^b]$. Then, also by the MVT, there exists some $d$ such that $e^d \in (e^a, e^b)$ and
$$e^{-d}f'(d) = \frac{1}{e^d} f'(\ln(e^d)) = \frac{f(\ln(e^b)) - f(\ln(e^a))}{e^b-e^a} = \frac{f(b) - f(a)}{e^b - e^a}.$$
Taking the quotient yields the desired result.
A: We recognize that the term $\frac{e^b - e^a}{b-a}$ resembles the form given by the Mean Value Theorem, namely $f'(c) = \frac{f(b) - f(a)}{b-a}$ for $a < c < b$.
Let $f(x) = e^x$ and note that $f$ is differentiable on $(a, b)$, continuous on $[a, b]$ and $f'(x) \ne 0$. Then by the Mean Value Theorem, there exists $c$ in $(a,b)$ such that
\begin{align*}
f'(c) &= \frac{f(b) - f(a)}{b-a} \\
&= \frac{e^b - e^a}{b-a}
\end{align*}
Dividing both sides by $f'(d) = e^d$, we obtain
\begin{align*}
\frac{f'(c)}{f'(d)} &= \frac{e^b - e^a}{b-a} \cdot \frac{1}{f'(d)} \\
&= \frac{e^b - e^a}{b-a} \cdot e^{-d}
\end{align*}
