How can I implicitly differentiate #(x-y)^3=0#?

2 Answers
Apr 20, 2015

If you insist on using implicit differentiation, you'd write:

#(x-y)^3=0#

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

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

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

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

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

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

Now we can say, either #x-y=0#, or we can divide by #x-y# to get #dy/dx = 1#

The difficulty is that the only solutions to #(x-y)^3=0# are #x-y=0#
so #y=x# and #dy/dx =1#. Why use implicit differentiation?

Apr 20, 2015

#dy/dx=1#

Rewrite it as #x^3 -3x^2y+3xy^2 -y^3#=0

Now differentiate term by term with respect to x,

#3x^2-6xy -3x^2dy/dx +3y^2 +6xydy/dx -3y^2 dy/dx#=0

#-3(x^2 -2xy+y^2)dy/dx= -3x^2 +6xy -3y^2#

#dy/dx=1#