Chain Rule proof with three functions

[PaulKraemer]

Hi,

I am stuck on a problem where I am asked to prove the following:

If z = k(y), y = f(u), and u = g(x), show that under suitable restrictions:

Dx z = (Dy z) (Du y) (Dx u)

Given the Chain Rule definition, it makes sense to me that this would be true, but when I try to prove it, I just end up going in circles.

If anyone could help me out, I'd really appreciate it.

Thanks in advance,
Paul

 

[JeffM]

Hi,

I am stuck on a problem where I am asked to prove the following:

If z = k(y), y = f(u), and u = g(x), show that under suitable restrictions:

Dx z = (Dy z) (Du y) (Dx u)

Given the Chain Rule definition, it makes sense to me that this would be true, but when I try to prove it, I just end up going in circles.

If anyone could help me out, I'd really appreciate it.

Thanks in advance,
Paul

There is probably an elegant proof by mathematical induction and the chain rule for 2 functions.

There is also a brute force method that probably mirrors how the pioneers of calculus did it back in the 17th century.

G(x + a) = (u + b).
So, b = G(x + a) - G(x).

F(u + b) = (y + c).
So, c = F(u + b) - F(u).

K(y + c) = (z + e).
So, e = K(y + c) - K(y).

[K(y + c) - K(y)] / a = (e / a) = (e / a) * (c / c) = (e / c) * (c / a) = (e / c) * (c / a) * (b / b) = (e / c) * (c / b) * (b / a).
So, [K(y + c) - K(y)] / a = {[K(y + c) - K(y)] / c} * {[F(u + b) - F(u)] / b} * {G(x + a) - G(x) / a}.

As a --> 0, b --> 0. As b --> 0, c --> 0.
So, under the requisite conditions, you just proceed to reduce the mess above to derivatives.

Galactus will come along eventually and tell you how to make this rigorous. I was trained as an historian so this is a handwaving argument, but it was probably good enough for Newton and Leibniz.

 

[PaulKraemer]

Thanks Jeff,

That helps alot!

Paul