Question

1. 0 / 5.55 points

differentiate.
$y = \frac{x^3}{3 - x^2}$
$y = \square$

Explanation:

Step1: Identify numerator and denominator

Let \( f(x) = x^3 \), \( g(x) = 3 - x^2 \)

Step2: Compute derivatives of f and g

\( f'(x) = 3x^2 \), \( g'(x) = -2x \)

Step3: Apply quotient rule

Quotient rule: \( y' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \)
Substitute values: \( y' = \frac{3x^2(3 - x^2) - x^3(-2x)}{(3 - x^2)^2} \)

Step4: Simplify numerator

\( 9x^2 - 3x^4 + 2x^4 = 9x^2 - x^4 = x^2(9 - x^2) \)

Answer:

\( \frac{x^2(9 - x^2)}{(3 - x^2)^2} \)