Differentiation & composite functions (2)

[kingwinner]

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

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

Thanks for explaining!

 

[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).

 

[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.

 

[kingwinner]

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).

OK, thanks!

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

Sorry my calculus is a bit rusty now...

 

[HallsofIvy]

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

 

[CompuChip]

OK, thanks!

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

Sorry my calculus is a bit rusty now...

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