Derivatives of Trig Functions — Question 8

Question 8 - Solution

We are given: $$f(x) = \frac{\sin(x)\cos(x)}{1 + \sin^2(x)}$$

Step 1: Rewrite using a double-angle identity.

Recall: $$\sin(x)\cos(x) = \tfrac{1}{2}\sin(2x)$$

So we rewrite the function: $$f(x) = \frac{\tfrac{1}{2}\sin(2x)}{1 + \sin^2(x)}$$

Step 2: Apply the quotient rule.

Let: $$u(x) = \tfrac{1}{2}\sin(2x), \qquad v(x) = 1 + \sin^2(x)$$

Differentiate each function:

$$u'(x) = \cos(2x)$$ $$v'(x) = 2\sin(x)\cos(x) = \sin(2x)$$

The quotient rule states: $$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$

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

Step 3: Simplify carefully.

Rewrite the squared sine: $$\sin^2(2x) = 1 - \cos^2(2x)$$

So the numerator becomes: $$\cos(2x)(1+\sin^2(x)) - \tfrac{1}{2}(1-\cos^2(2x))$$

Expand: $$= \cos(2x) + \cos(2x)\sin^2(x) - \tfrac{1}{2} + \tfrac{1}{2}\cos^2(2x)$$

Now use: $$\sin^2(x) = \tfrac{1-\cos(2x)}{2}$$

Substitute: $$\cos(2x)\sin^2(x) = \tfrac{1}{2}\cos(2x) - \tfrac{1}{2}\cos^2(2x)$$

Combine terms: $$= \tfrac{3}{2}\cos(2x) - \tfrac{1}{2}$$

Final Simplified Result: $$f'(x) = \frac{3\cos(2x) - 1}{2(1+\sin^2(x))^2}$$

Final Answer: $$\boxed{f'(x) = \frac{3\cos(2x) - 1}{2(1+\sin^2(x))^2}}$$