How do you use implicit differentiation to find dy/dx given #x^2-y^3=9#?

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Answer 1 · 2016-11-12T00:19:20

#dy/dx = (2x)/(3y^2)#

#x^2-y^3=9#

#2x-3y^2(dy/dx)=0#

#(dy/dx)(-3y^2)=-2x#

#(dy/dx)=(-2x)/(-3y^2)#

The answer is:
#dy/dx = (2x)/(3y^2)#

Answer 2 · 2016-11-12T00:21:58

Please see the explanation.

First term:

#d(x^2)/dx = 2x#

Second term:

#(d(-y^3))/dx# = -3y^2(dy/dx)#

Last term:

#(d(9))/dx = 0#

Put each term into its corresponding locaton:

#2x - 3y^2(dy/dx) = 0#

Solve for #dy/dx#:

#-3y^2(dy/dx) = -2x#

#dy/dx = (2x)/(3y^2)#