Question

How do you integrate `y=1/(-2+2x^2)` using the quotient rule?

Answer 1

I assume mean differentiate, not integrate, as the quotient rule is an differentiation tool!

`dy/dx = - (4x)/(-2+2x^2)^2`

Explanation:

You need to use the quotient rule; `d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`

So if `y = 1/(-2+2x^2)` we have;

`dy/dx = {(-2+2x^2)(d/dx1) - (1)(d/dx(-2+2x^2))}/(-2+2x^2)^2`
`:. dy/dx = {0 - (4x)}/(-2+2x^2)^2`
`:. dy/dx = - (4x)/(-2+2x^2)^2`