Find f'(x), and then evaluate f'left(π/6right).

By the basic trigonometric derivative rule, d/dx[sin(x)] = cos(x), so f'(x) = cos(x). Evaluating at x = π/6 gives f'(π/6) = cos(π/6) = frac√32.

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Derivative of sine and evaluation at a point

Let $f(x) = \sin(x)$ .

Find $f'(x)$ , and then evaluate $f'\left(\frac{\pi}{6}\right)$ .

Hint

Recall the basic trigonometric derivative rule for $\sin(x)$ , then substitute $x = \pi/6$ into the resulting expression and simplify using the known value of cosine at that angle.

Step-by-step solution

By the basic trigonometric derivative rule, $\frac{d}{dx}[\sin(x)] = \cos(x)$ , so $f'(x) = \cos(x)$ .

Evaluating at $x = \pi/6$ gives $f'(\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}$ .

Derivative of sine and evaluation at a point

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