# Second Derivative of a Composition of Functions

1. Jan 20, 2012

### skyline01

1. The problem statement, all variables and given/known data
Derive an expression for the composition of 2 functions.

2. Relevant equations

3. The attempt at a solution
I started with supposing that y(x) = f(g(x)). I know that dy/dx = df/dg * dg/dx (via the chain rule). Doing the derivative again, I started with the product rule:
d2y/dx2 = d/dx (df/dg * dg/dx) = df/dg * d2g/dx2 + dg/dx * d/dx (df/dg).

I don't know what d/dx (df/dg) means. I think I'm supposed to do the chain rule again on this part, but I'm not totally clear. Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 20, 2012

### SammyS

Staff Emeritus
Hello skyline01. Welcome to PF !

One suggestion is to to look at some specific examples, such as y(x) = sin(x5).

Then f(x) = sin(x), g(x) = x5 .

3. Jan 20, 2012

### skyline01

Thank you, SammyS! I tried the sample function you suggested, but I'm still not sure if I am expressing d/dx (df/dg) correctly. Here is how your sample function worked out:

y(x) = sin(x5)
So, f(x) = sin(x) and g(x) = x5.
Therefore,
dy/dx = cos(x5) 5x4
and
d2y/dx2 = cos(x5) 20x3 + 5x4 (-sin(x5)) 5x4.

Generalizing this to any composition of 2 functions, we have
d2y/dx2 = df/dg * d2g/dx2 + dg/dx * d2f/dg2.

Does this look right?

4. Jan 21, 2012

### SammyS

Staff Emeritus
Almost correct.

d2y/dx2 = df/dg * d2g/dx2 + (dg/dx)2 * d2f/dg2.

5. Jan 21, 2012

### skyline01

Yes, you are correct. Thank you for catching that. So,
d/dx(df/dg) = dg/dx * d2f/dg2.

I find it strange that in every calculus book I have ever read (including real analysis books), they always stop at the first derivative when discussing the chain rule.

6. Jan 21, 2012

### SammyS

Staff Emeritus
Yes, that is strange.

The notation can be confusing, even for 1st derivatives, but as you mentioned in your initial post, it gets quite a bit more confusing with higher order derivatives. That's why I suggested differentiating a concrete example.

Here's a link from Wikipedia: http://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives.

Have a good one !
SammyS