How do you implicitly differentiate #x^3 + y^3 = 9xy#?

1 Answer
Jul 9, 2015

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

Explanation:

When deriving implicitly you must remember to trea #y# as a function of #x# so that, for example, if you have:
#y^2# derived gives you: #2y*(dy)/(dx)#
In your case you have:
#3x^2+3y^2(dy)/(dx)=9y+9x(dy)/(dx)# where on the right I used the Product Rule:

Collecting #(dy)/(dx)# you get:

#(dy)/(dx)=(3x^2-9y)/(9x-3y^2)=(cancel(3)(x^2-3y))/(cancel(3)(3x-y^2))=(x^2-3y)/(3x-y^2)#