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

1 Answer
Jan 26, 2016

Please see the explanation section below.

Explanation:

Begin by rewriting. Add #y# to both sides to get rid of the duplicate terms.

#y^3=x^3+3xy#.

Note that the term #3xy# has a product of both variables in it. When we differentiate that term we will need the product rule.

Now, assuming that you are differentiating with respect to #x#, we get

#d/dx(y^3) = d/dx(x^3) + d/dx(3xy)#.

So #3y^2 dy/dx = 3x^2 + [3y+3x dy/dx]#

Solving for #dy/dx#, we have

#3y^2 dy/dx - 3x dy/dx = 3x^2 +3y#

#y^2 dy/dx - x dy/dx = x^2 +y#

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

So,

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

If you are differentiating with respect to #t#

we get:

#3y^2 dy/dt = 3x^2 dx/dt +3y dx/dt +3x dy/dt#.

So,

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