Derivatives of Trig Functions — Question 7

Question 7 - Solution

We are given: $$f(x) = \tan(x) \cdot \sec(x)$$

(a) Apply the Product Rule:

The product rule states: $$f'(x) = u'(x)v(x) + u(x)v'(x)$$

Let: $$u(x) = \tan(x), \quad v(x) = \sec(x)$$ $$u'(x) = \sec^2(x), \quad v'(x) = \sec(x)\tan(x)$$

Then: $$f'(x) = \sec^2(x) \cdot \sec(x) + \tan(x) \cdot \sec(x) \cdot \tan(x)$$ $$= \sec^3(x) + \sec(x)\tan^2(x)$$

(b) Factor the result:

Both terms have a factor of $\sec(x)$, so factor it out: $$f'(x) = \sec(x)\left(\sec^2(x) + \tan^2(x)\right)$$

Final Answer: $$\boxed{f'(x) = \sec(x)\left(\sec^2(x) + \tan^2(x)\right)}$$