How do you implicitly differentiate # x^3 + 3x^2 + y^3 = 8#?

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Answer 1 · 2016-04-11T22:11:08

You can do it like this:

#x^3+3x^(2)+y^(3)=8#

Differentiating both sides with respect to #x#:

#D(x^(3)+3x^(2)+y^(3))=D(8)#

#:.3x^(2)+6x+3y^(2)y'=0#

#:.3y^(2)y'=-(3x^(2)+6x)=-3(x^(2)+2x)#

#:.y'=(-cancel(3)(x^(2)+2x))/(cancel(3)y^2)#

#y'=((x^2+2x))/(y^(2))#