The converse is also true:
If f is a differentiable real-valued function on an open interval I and f'(x) is nonnegative (nonpositive) for all x in I, then f is increasing (decreasing) on I.
Here's a proof by contradiction:
Assume f is increasing on an open interval I and there exists a point x in I such that f'(x) < 0.
Since f is increasing, for any x1, x2 in I such that x1 < x2, we have f(x1) ≤ f(x2).
Since f'(x) < 0, we can choose an interval (a, x) such that f'(t) < 0 for all t in (a, x).
Consider a point y in (a, x) such that y > x.
By the Mean Value Theorem, there exists a point c in (a, x) such that:
f'(c) = (f(y) - f(x)) / (y - x)
Since f'(c) < 0, f(y) < f(x), which contradicts the fact that f is increasing.
Therefore, f'(x) cannot be negative for any x in I.
Assume f is decreasing on an open interval I and there exists a point x in I such that f'(x) > 0.
Since f is decreasing, for any x1, x2 in I such that x1 < x2, we have f(x1) ≥ f(x2).
Since f'(x) > 0, we can choose an interval (x, b) such that f'(t) > 0 for all t in (x, b).
Consider a point y in (x, b) such that y < x.
By the Mean Value Theorem, there exists a point c in (x, b) such that:
f'(c) = (f(x) - f(y)) / (x - y)
Since f'(c) > 0, f(x) < f(y), which contradicts the fact that f is decreasing.
Therefore, f'(x) cannot be positive for any x in I.