Question

For |x| = sqrt(x²) and using the Chain Rule, one can verify that d/dx |x| = x/|x|.

(a) If f(x) = |sin(x)|, find f ′(x).
(b) Where is f(x) not differentiable? Merely give the smallest positive value of x.

Answer 1

The derivative f'(x) of f(x) = |sin(x)| is cos(x) when sin(x) >= 0 and -cos(x) when sin(x) < 0. The function is not differentiable at the zeroes of sin(x), with the smallest positive value being x = π.

Step-by-step explanation:

To find f′(x) for f(x) = |sin(x)|, we need to consider two cases, as the absolute value affects the derivative differently depending on whether the sine function is positive or negative. When sin(x) ≥ 0, f′(x) will be cos(x), and when sin(x) < 0, f′(x) will be -cos(x). Therefore, we can write:

f′(x) = cos(x) if sin(x) ≥ 0

f′(x) = -cos(x) if sin(x) < 0

Function f(x) is not differentiable at points where sin(x) switches from positive to negative or vice versa, which occurs at the zeros of sin(x). The smallest positive value of x where this happens is x = π, since sin(0) = 0 and the next zero is at x = π.