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

1 Answer
RedRobin9688
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#

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