Let r(x) = cot(x). Find r'(x), and then evaluate r'left(π/4right).

Using the standard trigonometric derivative rule, d/dx[cot(x)] = -csc^2(x). So r'(x) = -csc^2(x). Now evaluate at x = π/4: since csc(π/4) = √2, we get csc^2(π/4) = 2. Therefore r'(π/4) = -2.

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Derivative of cotangent evaluated at a point

Let $r(x) = \cot(x)$ . Find $r'(x)$ , and then evaluate $r'\left(\frac{\pi}{4}\right)$ .

Hint

Recall the memorized derivative rule for $\cot(x)$ from the list of six trigonometric derivatives, then substitute $x = \pi/4$ and use $\csc(\pi/4) = \sqrt{2}$ .

Step-by-step solution

Using the standard trigonometric derivative rule, $\frac{d}{dx}[\cot(x)] = -\csc^2(x)$ . So $r'(x) = -\csc^2(x)$ .

Now evaluate at $x = \pi/4$ : since $\csc(\pi/4) = \sqrt{2}$ , we get $\csc^2(\pi/4) = 2$ . Therefore $r'(\pi/4) = -2$ .

Derivative of cotangent evaluated at a point

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