Differentiation Formulas — Question 10

Question 10 - Solution

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

Let: $$u = \sin(x), \quad v = 1 + \cos(x)$$

Then: $$f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}$$

Compute the derivatives: $$u' = \cos(x), \quad v' = -\sin(x)$$

Apply the quotient rule: $$f'(x) = \frac{(1 + \cos(x))\cos(x) - \sin(x)(-\sin(x))}{(1 + \cos(x))^2}$$

Simplify numerator: $$(1 + \cos(x))\cos(x) + \sin^2(x) = \cos(x) + \cos^2(x) + \sin^2(x)$$

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

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

Cancel common factor: $$f'(x) = \frac{1}{1 + \cos(x)}$$

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