What is the value of the implicit derivative of #y=y^2x^2# at #x=3#?

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1 Answer
mason m
Dec 13, 2015

#0,-2/27#

Explanation:

#d/dx[y=y^2x^2]#

#dy/dx=x^2d/dx[y^2]+y^2d/dx[x^2]#

#dy/dx=2x^2ydy/dx+2xy^2#

#dy/dx-2x^2ydy/dx=2xy^2#

#dy/dx(1-2x^2y)=2xy^2#

#dy/dx=(2xy^2)/(1-2x^2y)#

Now that we have the implicit derivative, we must find the point or points when #x=3#, so we can plug them in to find the value(s) of the implicit derivative.

We know that #x=3#, so:

#y=y^2x^2#
#y=9y^2#

With this:

#0=9y^2-y#
#0=y(9y-1)#

#y=0,1/9#

We now have to two points #(3,0)# and #(3,1/9)#.

#(3,0)rarr#

#dy/dx=(2(3)(0^2))/(1-2(3^2)(0))=0/1=0#

#(3,1/9)rarr#

#dy/dx=(2(3)(1/9)^2)/(1-2(3^2)(1/9))=(2/27)/(-1)=-2/27#

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