How do you implicitly differentiate $2 = e^{x y^{3} - x} - y^{2} x$ $2= e^(xy^3-x)-y^2x$ ?

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Answer 1 · 2018-06-05T17:21:14

$y'=(e^(xy^3-x(1-y^3)+y^2)/(3xy^2e^(xy^3-x)+2xy))$

Differentiating with respect to $x$
$0=e^(xy^3-x)(y^3+x*3y^2y'-1)-y^2-x*2y*y'$
this gives

$0=e^(xy^3-x)(y^3-1)-y^2+y'(3xy^2e^(xy^3-x)+2xy)$

How do you implicitly differentiate 2= e^(xy^3-x)-y^2x 2=exy3−x−y2x2= e^(xy^3-x)-y^2x ?