Derivatives of Trig Functions — Question 4

Question 4 - Solution

We are given: $$f(x) = \cot(x)\csc(x) + \cos^2(x)$$

(a) Differentiate each term separately.

Term 1: $\cot(x)\csc(x)$

Using the product rule: $$\frac{d}{dx}[\cot(x)\csc(x)] = \cot(x)(-\cot(x)\csc(x)) + \csc(x)(-\csc^2(x))$$ $$= -\cot^2(x)\csc(x) - \csc^3(x)$$

Term 2: $\cos^2(x)$

Using the chain rule: $$\frac{d}{dx}[\cos^2(x)] = 2\cos(x)(-\sin(x)) = -2\sin(x)\cos(x)$$

(b) Combine and simplify: $$f'(x) = -\cot^2(x)\csc(x) - \csc^3(x) - 2\sin(x)\cos(x)$$

Use the identities: $$\cot^2(x) = \csc^2(x) - 1 \qquad 2\sin(x)\cos(x) = \sin(2x)$$

$$-\cot^2(x)\csc(x) = -(\csc^2(x)-1)\csc(x) = -\csc^3(x) + \csc(x)$$

Substitute and combine like terms: $$f'(x) = \csc(x) - 2\csc^3(x) - \sin(2x)$$

Final Answer: $$\boxed{f'(x) = \csc(x) - 2\csc^3(x) - \sin(2x)}$$