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.