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

1 Answer
Aug 23, 2016

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

Explanation:

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)}.