Question

differentiate.
$y = \frac{x^3}{3 - x^2}$
$y = \square$
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differentiate.
$y = \frac{x + 7}{x^3 + x - 2}$
$y = \square$
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Explanation:

Step1: Identify quotient rule components

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

\( y'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}=\frac{3x^2(3-x^2)-x^3(-2x)}{(3-x^2)^2} \)

Step4: Simplify numerator

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

Step5: Identify second quotient components

Let \( h(x)=x+7 \), \( k(x)=x^3+x-2 \)

Step6: Compute derivatives of h and k

\( h'(x)=1 \), \( k'(x)=3x^2+1 \)

Step7: Apply quotient rule

\( y'=\frac{h'(x)k(x)-h(x)k'(x)}{[k(x)]^2}=\frac{1(x^3+x-2)-(x+7)(3x^2+1)}{(x^3+x-2)^2} \)

Step8: Simplify numerator

\( x^3+x-2-(3x^3+x+21x^2+7)=-2x^3-21x^2-9 \)

Answer:

For \( y=\frac{x^3}{3-x^2} \): \( \frac{9x^2 - x^4}{(3 - x^2)^2} \)
For \( y=\frac{x+7}{x^3+x-2} \): \( \frac{-2x^3 -21x^2 -9}{(x^3+x-2)^2} \)