find the derivative of the function.\n$f(x)=\frac{x^{3}-9}{x^{2}+2}$\n$f(x)=$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{3}-9$, $u^\prime=3x^{2}$, $v = x^{2}+2$, and $v^\prime = 2x$.

Step2: Apply quotient - rule

$f^\prime(x)=\frac{(3x^{2})(x^{2}+2)-(x^{3}-9)(2x)}{(x^{2}+2)^{2}}$.

Step3: Expand the numerator

Expand $(3x^{2})(x^{2}+2)=3x^{4}+6x^{2}$ and $(x^{3}-9)(2x)=2x^{4}-18x$. Then the numerator is $3x^{4}+6x^{2}-(2x^{4}-18x)=3x^{4}+6x^{2}-2x^{4}+18x=x^{4}+6x^{2}+18x$.

Answer:

$\frac{x^{4}+6x^{2}+18x}{(x^{2}+2)^{2}}$