Product and Quotient Rule — Question 2

Question 2 - Solution

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

Let $$u(x) = x^2 \sin(x), \quad v(x) = x + 1$$

Then using the **quotient rule**: $$f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}$$

We compute: $$u'(x) = \frac{d}{dx}[x^2 \sin(x)] = 2x \sin(x) + x^2 \cos(x) \quad \text{(product rule)}$$ $$v'(x) = \frac{d}{dx}[x + 1] = 1$$

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

Now expand and simplify the numerator: $$= \frac{(x + 1)(2x \sin(x) + x^2 \cos(x)) - x^2 \sin(x)}{(x + 1)^2}$$

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

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