Expert: Paul Klarreich - 10/24/2005

Question
use implicit differentiation to find dy/dx and then d^2y/dx^2

1) 2sqrt(y)=x-y
2) if x^3+y^3=16, find the value of d^2y/dx^2 at the point (2,2).

Answer
Hi, Amy,

Differentiating implicit functions is actually fairly easy, although sometimes we don't like the form of the answer.  You will have some EQUATION involving x and y, but the equation is NOT written like y = f(x), i.e. explicitly solved for y.  The process is this:

1. DON'T solve for y.  (Part of your grade will depend on how well you don't do this.)

2. Differentiate each SIDE of the equation, using Whatever rules apply -- product, quotient, etc. AND the chain rule, assuming that y is some function of x, even though we can't see it.

2A. When you see the 'y' appear, you use the chain rule like this:

   D(y) = dy/dx
   D(y^2) = 2y dy/dx
   D(sqrt(y)) = D(y^1/2) = 1/2y^-1/2 dy/dx

etc.  In other words, every term that involves y will produce an instance of dy/dx.  If it doesn't you made a mistake.

3. Solve algebraically for dy/dx.  (You can write y', but a little bit of sloppiness here can cost you.)
======== YOUR PROBLEMS =======
1) 2sqrt(y)=x-y

There are three terms:

The derivative of 2 sqrt(y) or 2y^1/2
would be 2(1/2)y^-1/2 dy/dx = (1/sqrt(y)) dy/dx

The derivative of x is 1

The derivative of y is dy/dx

Put them back into the equation:

(1/sqrt(y)) dy/dx = 1 + dy/dx

Solve:  (Bring all dy/dx terms on one side, all other terms to the other side.)

(1/sqrt(y)) dy/dx - dy/dx = 1
dy/dx[ 1/sqrt(y) - 1] = 1

dy/dx = 1/[ 1/sqrt(y) - 1]

That's it.  Yes, it involves y instead of x, but if you can't stand that, take some anti-nausea medication or something and accept it.


2) if x^3+y^3=16, find the value of d^2y/dx^2 at the point (2,2).  

This may take some more work.  When we have solved for dy/dx we will just differentiate again, and AT THE END, when dy/dx rears its ugly head again, substitute for it.

There are three terms:

D(x^3) = 3x^2
D(y^3) = 3y^2 dy/dx
D(16) = 0

3x^2 + 3y^2 dy/dx = 0

Solve:  dy/dx = -x^2/y^2

Differentiate again:  Note that the derivative of y^2 is 2y dy/dx.  We use the quotient rule:
            (y^2)(2x) - (x^2)(2y dy/dx)
d^2y/dx^2 =  ---------------------------
                       y^4

            2xy^2 - 2x^2 y (-x^2/y^2) <- value of dy/dx
d^2y/dx^2 =  -------------------------
                       y^4

            2xy^2 + 2x^4/y
d^2y/dx^2 =  ---------------
                  y^4

            2xy^3 + 2x^4
d^2y/dx^2 =  ---------------
                  y^5

OK. Now you can put  x = 2, y = 2  and get your answer.