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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: [tex]y = u^{n}[/tex] and u is a function of x (let's say) [tex]u = 3x - 2[/tex] therefore by composition y is a function of x.

By the definition of the derivative:

[tex]\frac{dy}{dx} = \lim_{\Delta x\to0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}}[/tex]

we can say that

[tex]\frac{dy}{dx} = \lim_{\Delta x\to0}{\frac{\Delta y}{\Delta x}}[/tex]

in the same way that

[tex]\frac{dy}{du} = \lim_{\Delta u\to0}{\frac{\Delta y}{\Delta u}}[/tex]

and

[tex]\frac{du}{dx} = \lim_{\Delta x\to0}{\frac{\Delta u}{\Delta x}}[/tex]

Then, using the equation:

[tex]\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}[/tex]

we can write

[tex]\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}[/tex]

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

[tex]\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}[/tex]

And so the chain rule.

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

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

Cheers!

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# Chain rule explanation

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