Product and Quotient Rule — Question 6

Question 6 - Solution

We are given: $$f(x) = \frac{x^2 e^{2x} \sin x}{1 + x^3}$$

Let $$u(x) = x^2 e^{2x} \sin x, \quad v(x) = 1 + x^3$$

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

Step 1: Differentiate $u(x)$

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

Apply the product rule: $$u'(x) = (x^2)' e^{2x} \sin x + x^2 (e^{2x})' \sin x + x^2 e^{2x} (\sin x)'$$

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

So: $$u'(x) = 2x e^{2x} \sin x + 2x^2 e^{2x} \sin x + x^2 e^{2x} \cos x$$

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

Step 2: Differentiate $v(x)$

$$v'(x) = 3x^2$$

Step 3: Apply the Quotient Rule

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

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

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