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

1 Answer
Jun 25, 2018

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