1. ## Implicit Differentiation

Use Implicit Differentiation:

(x^3) + (y^3) = 6xy

So I got:

(3x^2) + (3y^2)(dy/dx) = (0)(x*dy/dx) + (y)(1)

But the answer I got was wrong...any help?

2. Originally Posted by Dickson
Use Implicit Differentiation:

(x^3) + (y^3) = 6xy

So I got:

(3x^2) + (3y^2)(dy/dx) = (0)(x*dy/dx) + (y)(1)
Why "(0)(x*dy/dx)? Surely you aren't thinking "the derivative of 6 is 0" because you are differentiating y here, not 6. That first term should be 6x*dy/dx. And, for course, the second is 6y, not just y.

But the answer I got was wrong...any help?
In general, if c is a constant, (cf(x))'= cf'(x). If you really WANT to use the product rule, (cf(x))'= 0f(x)+ cf'(x)= cf'(x).

You might write, above, (6xy)'= (0)xy+ (6y)+ (6x)y'= 6xdy/dx+ 6y.

3. I surely was thinking 6 was 0...god. Anyway thanks for the help.

4. Well, 6 is close to 0!