How do you implicitly differentiate #6=lny/(e^x-x)-y^2#?

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Answer 1 · 2016-08-23T16:05:00

#y'={y(6+y^2)(e^x-1)}/{1-2y^2(e^x-x)}#.

Rewriting the given eqn. as, #(6+y^2)(e^x-x)=lny#.

Diff. ing both sides w.r.t. #x#, we get,

#(6+y^2)(e^x-x)'+(e^x-x)(6+y^2)'=(lny)'#.

#:. (6+y^2)(e^x-1)+(e^x-x)(0+2y.y')=1/y.y'#

#:. y(6+y^2)(e^x-1)+2y^2y'(e^x-x)=y'#

#:. y(6+y^2)(e^x-1)=y'{1-2y^2(e^x-x)}#

#:. y'={y(6+y^2)(e^x-1)}/{1-2y^2(e^x-x)}#.