How do you implicitly differentiate x^3 + y^3 = 9xyx3+y3=9xy?

1 Answer
Gió
Jul 9, 2015

I found: (dy)/(dx)=(x^2-3y)/(3x-y^2)dydx=x2−3y3x−y2

Explanation:

When deriving implicitly you must remember to trea yy as a function of xx so that, for example, if you have:
y^2y2 derived gives you: 2y*(dy)/(dx)2y⋅dydx
In your case you have:
3x^2+3y^2(dy)/(dx)=9y+9x(dy)/(dx)3x2+3y2dydx=9y+9xdydx where on the right I used the Product Rule:

Collecting (dy)/(dx)dydx 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)

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