Let $f$ be defined on $[a,b]$ s.t. $b$ is a minimum and $f$ is differentiable at $b$. Prove or disprove: $f'(b)\le 0.$
Solution 1:
As correctly pointed out by hamam_Abdallah in the comments,
$\ b\ $ is a minimum and so $\ \forall\ x\in\ [a,b),\ \frac{f(x)-f(b)}{x-b}\leq 0.$
By assumption, $\ f\ $ is differentiable at $\ b,\ $ and so $\ \lim_{x\to b^-} \frac{f(x)-f(b)}{x-b}\ $ exists.
But the limit of numbers $\ \leq 0\ $ must be $\ \leq\ 0,\ $ hence the result follows.
Solution 2:
Since $b$ is a minimum, you have $f(b)\leq f(x)$ for all $x\in [a,b]$. Thus for all $a\leq x<b$ we have $$\frac{f(x) - f(b)}{x-b} \leq 0. $$ Take the limit when $x\to b$, and you obtain $f'(b) \leq 0$.