Derivatives of Trig Functions — Question 3

Question 3

Let $$f(x) = \sin(x)\cos(x) + \tan(x)\sec(x)$$

• Find $f'(x)$.

• Simplify the expression.

Question 3 - Solution

We are given: $$f(x) = \sin(x)\cos(x) + \tan(x)\sec(x)$$

(a) Differentiate each term:

Term 1: $\sin(x)\cos(x)$

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

Term 2: $\tan(x)\sec(x)$

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

(b) Combine and simplify:

$$f'(x) = \cos^2(x) - \sin^2(x) + \sec^3(x) + \sec(x)\tan^2(x)$$

Use trigonometric identities: $$\cos^2(x) - \sin^2(x) = \cos(2x)$$ $$\tan^2(x) = \sec^2(x) - 1$$

$$\sec(x)\tan^2(x) = \sec^3(x) - \sec(x)$$

Substitute and combine like terms: $$f'(x) = \cos(2x) + 2\sec^3(x) - \sec(x)$$

Final Answer: $$\boxed{f'(x) = \cos(2x) + 2\sec^3(x) - \sec(x)}$$