How do you implicitly differentiate $2=e^ysiny-x^2y^3$ ?

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Answer 1 · 2016-04-04T10:53:25

$y'=(2xy^3)/[e^y*cos y+e^y*sin y-3x^2y^2]$

From the given equation

$2=e^y sin y-x^2y^3$

Differentiate both sides of the equation with respect to $x$

$d/dx(2)=d/dx(e^y sin y-x^2y^3)$

$0=e^y*d/dx(sin y)+sin y*d/dx(e^y)-[x^2*d/dx(y^3)+y^3*d/dx(x^2)]$

$0=e^y*(cos y)y'+sin y*(e^y)*y'-[x^2(3y^2)y'+(y^3)(2x)]$

$0=e^y*(cos y)y'+sin y*(e^y)*y'-x^2(3y^2)y'-(y^3)(2x)$

$0=[e^y*(cos y)+sin y*(e^y)-x^2(3y^2)]y'-(y^3)(2x)$

$(y^3)(2x)=[e^y*(cos y)+sin y*(e^y)-x^2(3y^2)]y'$

$y'=(2xy^3)/[e^y*cos y+e^y*sin y-3x^2y^2]$

God bless....I hope the explanation is useful.

How do you implicitly differentiate 2=e^ysiny-x^2y^3 2=e^ysiny-x^2y^3 ?