Question

differentiate the function.
g(x)=(x^3 - 8)cdot\frac{x^2 + 9}{x^2 - 9}
g(x)=square

Explanation:

Step1: Use product - rule

Let $u = x^{3}-8$, $v=\frac{x^{2}+9}{x^{2}-9}$. Product - rule: $G^\prime(x)=u^\prime v + uv^\prime$.
$u^\prime = 3x^{2}$.

Step2: Use quotient - rule for $v$

Quotient - rule: $v=\frac{f}{g}$, $v^\prime=\frac{f^\prime g - fg^\prime}{g^{2}}$, where $f = x^{2}+9$, $f^\prime = 2x$, $g=x^{2}-9$, $g^\prime = 2x$. Then $v^\prime=\frac{2x(x^{2}-9)-2x(x^{2}+9)}{(x^{2}-9)^{2}}=\frac{2x^{3}-18x - 2x^{3}-18x}{(x^{2}-9)^{2}}=\frac{-36x}{(x^{2}-9)^{2}}$.

Step3: Calculate $G^\prime(x)$

$G^\prime(x)=3x^{2}\cdot\frac{x^{2}+9}{x^{2}-9}+(x^{3}-8)\cdot\frac{-36x}{(x^{2}-9)^{2}}=\frac{3x^{2}(x^{2}+9)(x^{2}-9)-36x(x^{3}-8)}{(x^{2}-9)^{2}}=\frac{3x^{2}(x^{4}-81)-36x^{4}+288x}{(x^{2}-9)^{2}}=\frac{3x^{6}-243x^{2}-36x^{4}+288x}{(x^{2}-9)^{2}}$.

Answer:

$\frac{3x^{6}-36x^{4}-243x^{2}+288x}{(x^{2}-9)^{2}}$