How do you find #dy/dx# by implicit differentiation given #2x^2-3y^2=4#?

1 Answer
Feb 10, 2017

Take the derivative of both sides with respect to #x#, treating #y# as a function of #x#. Solve for #dy/dx#.

Explanation:

#2x^2-3y^2=4#

#=>"    "d/dx(2x^2-3y^2)"      "=d/dx(4)#

#=>d/dx(2x^2)-d/dx(3y^2)=0#

#=>"      "4x"     "-"   "6y dy/dx"   "=0"      "#(by chain rule)

#=>"                  "-6y dy/dx"   "=-4x#

#=>"                           "dy/dx"   "="  "("-"4x)/("-"6y)"   "="   "(2x)/(3y)#

Therefore, #dy/dx=(2x)/(3y)#.