Product and Quotient Rule — Question 10

Question 10 - Solution

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

This is a quotient of two functions: $$u(x) = x^2 e^x \sin(x), \quad v(x) = x^2 + 1$$

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

Step 1: Differentiate the numerator

The numerator is a product of three functions: $$u(x) = x^2 \cdot e^x \cdot \sin(x)$$

Using the product rule for three functions: $$u'(x) = (x^2)' e^x \sin(x) + x^2 (e^x)' \sin(x) + x^2 e^x (\sin(x))'$$

Compute derivatives: $$(x^2)' = 2x, \quad (e^x)' = e^x, \quad (\sin x)' = \cos x$$

Substitute: $$u'(x) = 2x e^x \sin(x) + x^2 e^x \sin(x) + x^2 e^x \cos(x)$$

Factor: $$u'(x) = e^x \left[ \sin(x)(2x + x^2) + x^2 \cos(x) \right]$$

Step 2: Differentiate the denominator

$$v'(x) = 2x$$

Step 3: Apply the quotient rule

$$f'(x) = \frac{ e^x \left[ \sin(x)(2x + x^2) + x^2 \cos(x) \right](x^2 + 1) - x^2 e^x \sin(x)(2x) }{(x^2 + 1)^2}$$

Step 4: Simplify

Factor out $e^x$: $$f'(x) = \frac{ e^x \left\{ (x^2 + 1)\left[ \sin(x)(2x + x^2) + x^2 \cos(x) \right] - 2x^3 \sin(x) \right\} }{(x^2 + 1)^2}$$

This is a fully factored and simplified form.

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