How do you implicitly differentiate # y^3+x^2-y/x^2-3y^2#?

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Answer 1 · 2018-06-24T05:58:46

The answer is #=(2(x^4+y))/(x(3x^2y^2-1-6x^2y))#

Let

#f(x,y)=y^3+x^2-y/x^2-3y^2#

Then the partial derivatives are

#(delf)/(delx)=2x+2y/x^3=2(x+y/x^3)=2(x^4+y)/x^3#

#(delf)/(dely)=3y^2-1/x^2-6y=(3x^2y^2-1-6x^2y)/x^2#

Therefore,

#dy/dx=-((delf)/(delx))/((delf)/(dely))#

#=(2(x^4+y)/x^3)/((3x^2y^2-1-6x^2y)/x^2)#

#=(2(x^4+y))/(x(3x^2y^2-1-6x^2y))#