Question

video example 4 let (y = \frac{x^{2}+x - 6}{x^{3}+8}). use the quotient rule to find (y). solution (y=\frac{(x^{3}+8)\frac{d}{dx}(x^{2}+x - 6)-(x^{2}+x - 6)\frac{d}{dx}(x^{3}+8)}{(x^{3}+8)^{2}}=\frac{(x^{3}+8)(2x + 1)-(x^{2}+x - 6)(3x^{2})}{(x^{3}+8)^{2}}=\frac{(x^{3}+8)(2x + 1)-(3x^{4}+3x^{3}-18x^{2})}{(x^{3}+8)^{2}}=\frac{2x^{4}-2x^{3}+18x^{2}+16x + 8}{(x^{3}+8)^{2}}).

Explanation:

Step1: Recall quotient rule

The quotient rule for $y = \frac{u}{v}$ is $y'=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$. Here $u = x^{2}+x - 6$ and $v=x^{3}+8$.

Step2: Differentiate $u$ and $v$

$\frac{du}{dx}=2x + 1$ and $\frac{dv}{dx}=3x^{2}$.

Step3: Substitute into quotient - rule formula

$y'=\frac{(x^{3}+8)(2x + 1)-(x^{2}+x - 6)(3x^{2})}{(x^{3}+8)^{2}}$.

Step4: Expand numerator

Expand $(x^{3}+8)(2x + 1)=2x^{4}+x^{3}+16x + 8$ and $(x^{2}+x - 6)(3x^{2})=3x^{4}+3x^{3}-18x^{2}$. Then $y'=\frac{2x^{4}+x^{3}+16x + 8-(3x^{4}+3x^{3}-18x^{2})}{(x^{3}+8)^{2}}$.

Step5: Simplify numerator

$y'=\frac{2x^{4}+x^{3}+16x + 8 - 3x^{4}-3x^{3}+18x^{2}}{(x^{3}+8)^{2}}=\frac{-x^{4}-2x^{3}+18x^{2}+16x + 8}{(x^{3}+8)^{2}}$.

Answer:

$y'=\frac{-x^{4}-2x^{3}+18x^{2}+16x + 8}{(x^{3}+8)^{2}}$