How do you implicitly differentiate #-y^2=e^(2x-4y)-3x #?

1 Answer
Jun 29, 2016

#:.dy/dx={3-2e^(2x-4y)}/[4{2y-e^(2x-4y)}].#

OR

#.dy/dx={3-6x+2y^2}/{4(2y-3x+y^2}.#

Explanation:

We write the eqn. of the curve as #: 3x-y^2=e^(2x-4y)..........(1)#

Diff.ing both sides w.r.t. x, #d/dx(3x-4y^2)=d/dxe^(2x-4y).#

#:. d/dx3x-d/dx4y^2=e^(2x-4y)*d/dx(2x-4y).#
#:.3-4d/dy(y^2)dy/dx=e^(2x-4y)(d/dx2x-d/dx4y)#......[Chain Rule]
#:.3-4(2y)dy/dx=e^(2x-4y){2-4dy/dx}.#
#:.3-2e^(2x-4y)=8ydy/dx-4e^(2x-4y)dy/dx=4{2y-e^(2x-4y)}dy/dx.#
#:.dy/dx={3-2e^(2x-4y)}/[4{2y-e^(2x-4y)}].#

This is fairly a good answer, but, we can further simplify it, replacing #e^(2x-4y),# by #3x-y^2# [bcz. of #(1)#], as under :-

#.dy/dx={3-2(3x-y^2)}/[4{2y-(3x-y^2)}]={3-6x+2y^2}/{4(2y-3x+y^2}.#