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I learned from a textbook that if $f(x)$ can be expressed as a function of $u(x)$, for example $$f(x) = u(x)^3,$$ and if $\delta f$, $\delta u$, $\delta x$ are small finite quantities, then $${\delta f \over \delta x} = {\delta f \over \delta u} {\delta u \over \delta x}.$$ As the quantities become infinitesimally small, we get $$\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}.$$ The above definitions seems to suggest the the quantities ${\delta u}$ and $du$ have to cancel in the expressions.

I expanded the above definition using the gradient definition of a derivative into $${df \over dx}=\lim_{\delta u \to 0} {f(u+\delta u)-f(u) \over \delta u} \times \lim_{\delta x \to 0} {u(x+\delta x)-u(x) \over \delta x}$$

Does this imply that the $\delta u$ and $u(x+\delta x)-u(x)$ cancel each other?If that is so, why are they equal? It does not seem intuitive to me that they equal.

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  • $\begingroup$ How do you define $\delta u$ ? $\endgroup$
    – user65203
    Jun 13, 2018 at 7:28
  • $\begingroup$ @TaeNyFan Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Aug 4, 2018 at 21:03

3 Answers 3

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$${\delta f \over \delta x} = {\delta f \over \delta u} {\delta u \over \delta x}$$ is an ordinary relation between fractions and the $\delta u$ truly cancel each other as they are the same number.

When going to the limit, by the definition of the derivatives and the product rule,

$$\frac{df}{dx}=\lim_{\delta x\to0}{\delta f \over \delta x} = \lim_{\delta x\to0}{\delta f \over \delta u} {\delta u \over \delta x}=\lim_{\color{blue}{\delta x\to0}}{\delta f \over \delta u}\lim_{\delta x\to0} {\delta u \over \delta x}=\lim_{\color{blue}{\delta u\to0}}{\delta f \over \delta u}\lim_{\delta x\to0} {\delta u \over \delta x}=\frac{df}{du}\frac{du}{dx}.$$

The subsitution of $\delta x$ for $\delta u$ (in blue) is possible because $\delta u$ and $\delta x$ go to zero simultaneously.

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  • $\begingroup$ This argument is fine as long as $\delta u$ doesn't vanish. If it does (e.g. $u$ is constant), this simple proof of the chain rule fails although the rule remains true. Is there a neat way to handle this? $\endgroup$ Jun 13, 2018 at 17:13
  • $\begingroup$ @JoonasIlmavirta: the rule remains true, but the rule $df/dx=(-df/du+\pi r^2\cos x)\,du/dx$ is true as well. There is no real point keeping the theorem in this case. $\endgroup$
    – user65203
    Jun 13, 2018 at 18:49
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We can give such kind of not rigorous but intuitive explanation of chain rule as follow for

$$f(x) = f(u(x))$$

then

$$\Delta f\approx f'(u(x))\Delta u$$

$$\Delta u\approx u'(x)\Delta x$$

thus

$$\Delta f\approx f'(u(x))u'(x)\Delta x\implies\frac {\Delta f}{\Delta x}\approx f'(u(x))u'(x)\implies \frac{d f}{dx}=f'(u(x))u'(x)=\frac{df}{du}\frac{du}{dx}$$

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By definition, you have

\begin{align} \delta f(u) &= f(u + \delta u) - f(u) \\ \delta u(x) &= u(x + \delta x) - u(x) \end{align}

The fact that the ratio cancels is immediately obvious here

$$ \frac{df}{du}\frac{du}{dx} = \lim_{\delta u \to 0}\frac{f(u + \delta u) - f(u)}{u(x+\delta x) - u(x)}\cdot \lim_{\delta x \to 0} \frac{u(x + \delta x) - u(x)}{\delta x} = \lim_{\delta x\to 0} \frac{f(u+\delta u)-f(u)}{\delta x} $$

The part you should care more about is going back from $u$ to $x$

\begin{align} f(u(x)) &= f(x) \\ f(u(x) + \delta u(x)) &= f\big(u(x+\delta x)\big) = f(x + \delta x) \end{align}

Thus

$$ \lim_{\delta x\to 0} \frac{f(u+\delta u)-f(u)}{\delta x} = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x} = \frac{df}{dx} $$

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