# Chain rule explanation

1. Jan 7, 2012

### cupu

Hello,

Looking through a book on calculus I found the following explanation for the chain rule and I have one unclear thing that I'd like to ask for help on.

The canonical example is used, y is a function of u: $$y = u^{n}$$ and u is a function of x (let's say) $$u = 3x - 2$$ therefore by composition y is a function of x.

By the definition of the derivative:

$$\frac{dy}{dx} = \lim_{\Delta x\to0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}}$$
we can say that
$$\frac{dy}{dx} = \lim_{\Delta x\to0}{\frac{\Delta y}{\Delta x}}$$

in the same way that
$$\frac{dy}{du} = \lim_{\Delta u\to0}{\frac{\Delta y}{\Delta u}}$$
and
$$\frac{du}{dx} = \lim_{\Delta x\to0}{\frac{\Delta u}{\Delta x}}$$

Then, using the equation:
$$\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}$$

we can write

$$\frac{dy}{dx} = \lim_{\Delta x\to0}\frac{\Delta y}{\Delta x} = \lim_{\Delta x\to0}(\frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}) = \lim_{\Delta x\to0}\frac{\Delta y}{\Delta u}\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x}$$

then it is said that "However because u is a function of x, $$\Delta x \rightarrow 0$$ exactly when $$\Delta u \rightarrow 0$$ so

$$\lim_{\Delta x\to0}\frac{\Delta y}{\Delta u} \lim_{\Delta x\to0}\frac{\Delta u}{\Delta x} = \lim_{\Delta u\to0}\frac{\Delta y}{\Delta u}\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x} = \frac{dy}{du} \cdot \frac{du}{dx}$$

And so the chain rule.

The part in red is the one I couldn't understand, specifically why does $$\Delta x \rightarrow 0$$ exactly when $$\Delta u \rightarrow 0$$ ?

Sorry for the lengthy post and thanks in advance for any help.

Cheers!

2. Jan 7, 2012

### micromass

"Proofs" like this want to make me rip my hair out.

3. Jan 7, 2012

4. Jan 7, 2012

### sahil_time

Yes, "u" is a function of "x". This implies that u is dependent on x and x is
an independent variable. Now since u depends on x, any change in x
produces a change in u. But if x does not change, there will be no
"CHANGE" in the dependent variable u.
:)

5. Jan 7, 2012

### sahil_time

Or,
Δu = Δf(x)
Δu = [3*(x+Δx) - 2] - [3*x - 2]
Δu = 3*Δx
now if Δx→0
then Δu→0.

Hope this helps.
:)

6. Jan 7, 2012

### Fredrik

Staff Emeritus
That particular $\Delta u$ is defined by $\Delta u=u(x+\Delta x)-u(x)$. Clearly $\Delta x=0$ implies $\Delta u=0$. The converse implication is explained here. However, I agree with micromass. This isn't the way to prove the chain rule. The problem is that $\Delta u$ is defined as I said when it appears in a numerator but not when it appears in a denominator.

The Wikipedia page has some short proofs, but I think I still prefer to prove it directly from the definitions, without the tricks that are used in the Wikipedia proofs. Unfortunately such a proof is about 2-3 pages long if all the details are explained.

Last edited: Jan 7, 2012
7. Jan 7, 2012

### cupu

@micromass
@spamiam
Thanks for the replies, I've seen other proofs which I've been able to workout on paper and those make sense, I just got really hung up on this specific 'proof' because I couldn't make sense of the red part.

@sahil_time
@Fredrik
Thank you! I though that the 'proof' presented in the book was for the general case and it didn't make sense so I didn't bother to work it out on paper, I can see now it was strictly (well .. more or less) related to the example provided; sorry for wasting your time, it's clear now!

So with that being said, my problem is resolved, the input was great, thanks a lot!

Cheers!