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,

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

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

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

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

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

How do you implicitly differentiate 6=lny/(e^x-x)-y^26=lny/(e^x-x)-y^2?