How do you find y'' by implicit differentiation of x^3+y^3=1 ?

1 Answer
Sep 6, 2014

Differentiate one step at a time, then when you have a (d^2y)/dx^2 term (i.e. y'') or can easily get one, rearrange the equation.

Implicit differentiation is remarkably similar to "regular" differentiation. We just need to treat any term with a y in it slightly differently.

First, we differentiate both sides of the equation:
d/dx(x^3+y^3) = d/dx(1)

By the addition rule:
d/dx(x^3+y^3) = d/dx(x^3)+d/dx(y^3)

We know that d/dxx^3 = 3x^2 and d/dx1=0, so
3x^2+d/dx(y^3)=0

Now we use the chain rule to implicitly find d/dx(y^3):
Let u = y^3
Then (du)/dy = 3y^2
and since (du)/dy*dy/dx = (du)/dx
we have d/dx(y^3) = 3y^2*dy/dx.

So we have 3x^2+3y^2*dy/dx = 0, and we can rearrange our equation to get dy/dx = (-3x^2)/(3y^2).

From here, the next step is to again differentiate both sides, this time using the quotient rule:
(d^2y)/dx^2=d/dx(-(3x^2)/(3y^2)) = -d/dx((3x^2)/(3y^2))
=(3y^2*d/dx(3x^2)-3x^2*d/dx(3y^2))/(3y^2)^2
=-(6x3y^2 - 3x^2*d/dx(3y^2))/(9y^4)

To find d/dx(3y^2), we again implicitly differentiate to get 6y*dy/dx

And so:
(d^2y)/dx^2 = -(18xy^2 - 18x^2y*dy/dx)/(9y^4)

Finally, recall that dy/dx = (-3x^2)/(3y^2)

So: (d^2y)/dx^2 = -(18xy^2 - 18x^2y*(-3x^2)/(3y^2))/(9y^4).