Question

Find dy/dx? (2x-3y^2)^4 = 6x+9y

Answer 1

`dy/dx=(8(2x-3y^2)^3-6)/(9+24y(2x-3y^2)^3`

Explanation:

1. Use substitution

Let `u=2x-3y^2` therefore `(du)/dx=2-6ydy/dx`

This results in `u^4=6x+9y`

1. Differentiate

`4u^3*(du)/dx=6+9dy/dx`

1. Back-sub original variables

`4(2x-3y^2)^3*(2-6ydy/dx)=6+9dy/dx`

1. Expand

`8(2x-3y^2)^3-24ydy/dx(2x-3y^2)^3=6+9dy/dx`

1. Separate terms involving `dy/dx`

`8(2x-3y^2)^3-6=9dy/dx+24y(2x-3y^2)^3dy/dx`

1. Combine like terms

`8(2x-3y^2)^3-6=(9+24y(2x-3y^2)^3)dy/dx`

1. Isolate `dy/dx`

`(8(2x-3y^2)^3-6)/(9+24y(2x-3y^2)^3)=dy/dx`