Derivatives of Trig Functions — Question 1

Question 1 - Solution

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

(a) Differentiate each term:

We apply the product rule to the first term $\sin(x)\cos(x)$, and recall the derivative of $\tan(x)$ is $\sec^2(x)$:

$$\frac{d}{dx}[\sin(x)\cos(x)] = \cos(x)\cos(x) + \sin(x)(- \sin(x)) = \cos^2(x) - \sin^2(x)$$

So the derivative of the entire function is: $$f'(x) = \cos^2(x) - \sin^2(x) + \sec^2(x)$$

(b) Simplify the result:

Note that: $$\cos^2(x) - \sin^2(x) = \cos(2x)$$

Thus: $$\boxed{f'(x) = \cos(2x) + \sec^2(x)}$$