Question

Find the deriative dy/dx for y=x^2-2x/x^3+3

Answer 1

Answer:

$\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})$

Step-by-step explanation:

So we want to find the derivative of the rational equation:

$y=\frac{x^2-2x}{x^3+3}$

First, recall the quotient rule:

$(\frac{f}{g})'=\frac{f'g-fg'}{g^2}$

Let f be x^2-2x and let g be x^3+3.

Calculate the derivatives of each:

$f=x^2-2x\\f'=2x-2$

$g=x^3+3\\g=3x^2$

So:

$\frac{dy}{dx}=(\frac{x^2-2x}{x^3+3})'$

Use the above format:

$\frac{dy}{dx}=\frac{f'g-fg'}{g^2}\\\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})$

And that's our answer :)

(If you want to, you can also expand. However, no terms will be canceled.)