Answer:
[tex]\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})[/tex]
Step-by-step explanation:
So we want to find the derivative of the rational equation:
[tex]y=\frac{x^2-2x}{x^3+3}[/tex]
First, recall the quotient rule:
[tex](\frac{f}{g})'=\frac{f'g-fg'}{g^2}[/tex]
Let f be x^2-2x and let g be x^3+3.
Calculate the derivatives of each:
[tex]f=x^2-2x\\f'=2x-2[/tex]
[tex]g=x^3+3\\g=3x^2[/tex]
So:
[tex]\frac{dy}{dx}=(\frac{x^2-2x}{x^3+3})'[/tex]
Use the above format:
[tex]\frac{dy}{dx}=\frac{f'g-fg'}{g^2}\\\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})[/tex]
And that's our answer :)
(If you want to, you can also expand. However, no terms will be canceled.)
