Second Derivative of a Composition of Functions

[skyline01]

Homework Statement

Derive an expression for the composition of 2 functions.

Homework Equations

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!

 

[SammyS]

Homework Statement

Derive an expression for the composition of 2 functions.

Homework Equations

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!

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 .

 

[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?

 

[SammyS]

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?

Almost correct.

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

 

[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.

 

[SammyS]

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