$f$ strictly increasing does not imply $f'>0$ We know that a function $f: [a,b] \to \mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f'&gt;0 \mbox{ on} (a,b)$ , f is strictly increasing on $[a,b]$. Is there any counterexample that shows the converse fails? 

I have been trying to come up with simple examples but they all involve functions that are discontinuous or has derivative $f'=0$ which does not agree with the hypothesis hmmm
Consider $f(x)=x^3$ on $[-1,1]$. It is strictly increasing, but has zero derivative at $0$.
Definition of Strictly Increasing: 
if $f(x)$ is strictly increasing then, 

$\forall x\forall y : y&gt;x, f(y)&gt; f(x) $

Definition of Monotonically increasing: 
if $f(x)$ is monotonically increasing then, 
$\forall x\forall y : y\geq x, f(y)\geq f(x) $

Definition of Derivative: 
$f'(x) = \lim_{∆x\to 0} \frac{f(x+∆x)-f(x)}{∆x} $

Postulate: $f(x)$ is a Strictly Increasing function 
$\iff$ $f'(x) &gt;0$ 

Proof: WLOG, let $ ∆x\overset{\underset{\mathrm{def}}{}}{=} y-x$ 

Thus, $y = x + ∆x $ 

$f'(x) = \lim_{(y-x) \to 0} \frac{f(y)-f(x)}{y-x} $

$\because y &gt; x, ∆x &gt;0 $ and $ y-x &gt; 0 $ 

also, $\because f(y) &gt; f(x), f(y) - f(x) &gt; 0 $ 

Thus $ \lim_{∆x\to 0} \frac{f(x+∆x)-f(x)}{∆x} &gt; 0 $

$ \therefore f'(x) &gt; 0$ QED

Using the same logic, $f(x)$  is monotonic $\iff f'(x) \geq 0$