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

Staff Emeritus
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.