Answer
If $f(x)$ has a local minimum at $c$ and $f'(c)$ exists, then $f'(c) = 0$
Work Step by Step
Suppose $f(x)$ has a local minimum at $c$ and $f'(c)$ exists.
According to Exercise 78, $g(x) = -f(x)$ has a local maximum at $c$
Also, since $f'(c)$ exists then $g'(c) = -f'(c)$ exists
According to Fermat's Theorem, $g'(c) = 0$
Then:
$-f(x) = g(x)$
$-f'(x) = g'(x)$
$f'(x) = -g'(x)$
$f'(c) = -g'(c) = -(0) = 0$
This shows that if $f(x)$ has a local minimum at $c$ and $f'(c)$ exists, then $f'(c) = 0$
