Let $f : [a,b] \to R $ be a differentiable function such that $f(x)\neq0$ for all $x\in [a,b]$. Then find the value of $\frac{f'(c)}{f(c)}$ for some $c\in(a,b)$
My attempt
I used from Mean Value Theorem which gives for some $c\in(a,b)$ $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ or $$\frac{f'(c)}{f(b)-f(a)}=\frac{1}{b-a}$$Can we consider $f(b)-f(a)=f(c)$? If so, what will be the value of $a$ and $b$ in terms of $c$? I may be wrong. I will be happy if someone can point out it.
Since $f(x)\ne 0$ for all $x$ we have $f(x)>0$ for all $x$ or $f(x)<0$ for all $x$.
Say $f(x)>0$ for all $x$. Probably you should look at the function $g(x) =\ln f(x)$ and use here MVT whatever you want to do.
By MVT there is a $c\in (a,b)$ such that $$g'(c) = {g(b)-g(a)\over b-a}$$ so
$${f'(c)\over f(c)} = {1\over b-a}\ln {f(b)\over f(a)}$$