# Differentiation & composite functions (2)

1. Oct 23, 2008

### kingwinner

Suppose we are given that
d
--F(y) = f(y)
dy

Then is it true that
dF(h(u))
-------- = f(h(u)) dh(u)/du ?
du
Why or why not? In particular, I don't understand how to get the red part...

Thanks for explaining!

Last edited: Oct 23, 2008
2. Oct 23, 2008

### CompuChip

It is basically just application of the chain rule.
Calling y = h(u),

$$\frac{dF(y)}{du} = \frac{dF(y)}{dy} \times \frac{dy}{du}$$
so if you call dF(y)/dy = f(y) then you have your identity (just replace y with its definition h(u) again).

3. Oct 23, 2008

### Ben Niehoff

Try this: http://en.wikipedia.org/wiki/Chain_rule

There is a proof about halfway down the page.

If that proof is confusing, then I should think your calculus textbook ought to provide a proof also, which may be easier to follow.

Or you can try to prove it yourself using the limit definition of the derivative.

4. Oct 23, 2008

### kingwinner

OK, thanks!

But can we replace y by h(u) and vice versa that freely?

Sorry my calculus is a bit rusty now...

5. Oct 24, 2008

### HallsofIvy

You can replace anything by anything that is equal to it!

6. Oct 24, 2008

### CompuChip

I defined y to be h(u). So, as HallsOfIvy points out, anywhere you see one you can replace it by the other.