Answer to Question #749 in Calculus for Luke

Question

Using the definition of the derivative, determine f ′ (0) for
f (x) = cos x, x < 0
1 x ≥ 0.
Is f ′ continuous at 0?

Answer 1

For x < 0 :f'(x) = (cos x)' = -sin x.
When x tends to zero: x->0 f'(x) = -sin 0 = 0.

For x ≥ 0:
f'(x) = (1)' = 0

As f'(x) at x =0 equals to 0 to the both sides from the point, it's continuous.