Derivatives of Trig Functions — Question 6

Question 6 - Solution

We are given: $$f(x) = \frac{\sin(x)}{1 + \cos(x)}$$

(a) Use the Quotient Rule:

The quotient rule is: $$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$

Let: $$g(x) = \sin(x), \quad h(x) = 1 + \cos(x)$$ $$g'(x) = \cos(x), \quad h'(x) = -\sin(x)$$

Then: $$f'(x) = \frac{\cos(x)(1 + \cos(x)) - \sin(x)(- \sin(x))}{(1 + \cos(x))^2}$$

$$= \frac{\cos(x)(1 + \cos(x)) + \sin^2(x)}{(1 + \cos(x))^2}$$

(b) Simplify the numerator:

First expand: $$\cos(x)(1 + \cos(x)) = \cos(x) + \cos^2(x)$$

So the full numerator is: $$\cos(x) + \cos^2(x) + \sin^2(x)$$

Use the identity: $$\cos^2(x) + \sin^2(x) = 1$$

So the numerator becomes: $$\cos(x) + 1$$

Therefore, $$f'(x) = \frac{\cos(x) + 1}{(1 + \cos(x))^2}$$

Simplify: $$f'(x) = \frac{1}{1 + \cos(x)}$$

Final Answer: $$\boxed{f'(x) = \frac{1}{1 + \cos(x)}}$$