In Leibniz notation of the chain rule, $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

Where $y\left ( u\left ( x \right ) \right )$ is a composite function of x.

I understand that the du's don't simply cancel out because $\frac{dy}{du}$ and $\frac{du}{dx}$ are defined as specific limits making the numerator and denominator infinitesimals and thus making the whole thing indeterminate and inoperable on.

But applying the definition of a derivative, we can express the above like so:

$$\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta u\to 0} \frac{\Delta y}{\Delta u} \cdot \lim_{\Delta x\to 0} \frac{\Delta u}{\Delta x}$$

At this point we can't use the Product Law of Limits to combine the two limits on the right.

But if we consider a coordinate system u vs x, doesn't $\Delta u \rightarrow 0$ when $\Delta x \rightarrow 0$ ? And if so, then whenever $\Delta u \rightarrow 0$ we necessarily have $\Delta x \rightarrow 0$.

Then can't we rewrite the above limit equation as:

$$\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\to 0} \frac{\Delta y}{\Delta u} \cdot \lim_{\Delta x\to 0} \frac{\Delta u}{\Delta x}$$

And then can't we use the Product Law of Limits to say:

$$\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\to 0} \left ( \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x} \right )$$

And since $\frac{\Delta y}{\Delta u}$ and $\frac{\Delta u}{\Delta x}$ within the quantity who's limit is being taken are no longer "quotients" infinitesimals, Δu's can cancel, leaving us with:

$$\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} $$

Which in Leibniz notations looks like:

$$\frac{dy}{dx} = \frac{dy}{dx}$$

Q.E.D (Since we manipulated the right hand side of the equation into looking the same as the left hand side).

  • $\begingroup$ Why do you make this assumption "if we consider a coordinate system u vs x, doesn't $\Delta u \rightarrow 0$ when $\Delta x \rightarrow 0$"?. May need to consider a practical example where the relation between u and x is inversely proportional. $\endgroup$ – NoChance Oct 14 '15 at 20:36
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    $\begingroup$ I'll point out here that the bit about not treating $dy/dx$, etc. as ordinary fractions in deriving $dy/dx = (dy/du)(du/dx)$ is not just mathematicians being pedantic. It's false in, for example, higher dimensions: the corresponding chain rule for a function $f(t) = g(x(t), y(t))$ is $\frac{df}{dt} = \frac{\partial g}{\partial x}\frac{dx}{dt} + \frac{\partial g}{\partial y} \frac{dy}{dt}$. $\endgroup$ – anomaly Oct 14 '15 at 20:36
  • $\begingroup$ NoChance - The relationship between u and x may be governed by inverse proportionality but as far as I can tell the relationship between $\Delta u$ and $\Delta x$ is always proportional. $\endgroup$ – Vahram Oct 14 '15 at 21:41

If $\Delta u=0$ then you have a $0$ in a denominator.

That's not a problem if it happens only when $|\Delta x|>0.000000000001$ since the limit depends only on what happens when $|\Delta x|$ is less than that. And similarly with any other positive number in place of $0.000000000001$.

But now suppose it happens when $|\Delta x|=0.000000000001$ and again when $|\Delta x| =0.000000000001^2$ and again when $|\Delta x|=0.000000000001^3$ and so on, ad infinitum. Then it's a difficulty to be addressed. And what if $\Delta u=0$ for all $\Delta x$ between $\pm 0.001$? Then your proof clearly won't work.

Hence one writes $$ \frac{\Delta y}{\Delta x} = \left.\begin{cases} \Delta y/\Delta u & \text{if } \Delta u\ne0, \\[6pt] dy/du & \text{if }\Delta u = 0, \end{cases} \right\} \cdot \frac{\Delta u}{\Delta x} $$ and one goes on from there. The $\displaystyle \left\{ \begin{array}{c} \text{factor in} \\ \text{braces} \end{array} \right\}$ approaches $dy/du$ as $\Delta x\to 0$ and the second factor, $\dfrac{\Delta u}{\Delta x}$, approaches $du/dx$.

  • $\begingroup$ You're talking about making that adjustment here: $$\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\to 0} \left ( \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x} \right )$$ Right? $\endgroup$ – Vahram Oct 14 '15 at 20:48
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    $\begingroup$ Yes. $\vphantom{\dfrac{\displaystyle\int}{\displaystyle\int}}$ $\displaystyle \lim_{\Delta x\to0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\to0} \left( \left.\begin{cases} \Delta y/\Delta u & \text{if } \Delta u\ne0, \\[6pt] dy/du & \text{if }\Delta u = 0, \end{cases} \right\} \cdot \frac{\Delta u}{\Delta x} \right)$ $\displaystyle= \left(\lim_{\Delta x\to0} \left.\begin{cases} \Delta y/\Delta u & \text{if } \Delta u\ne0, \\[6pt] dy/du & \text{if }\Delta u = 0, \end{cases} \right\} \right) \cdot \left( \lim_{\Delta x\to0} \frac{\Delta u}{\Delta x} \right)$. ${} \qquad {}$ $\endgroup$ – Michael Hardy Oct 14 '15 at 20:56
  • $\begingroup$ Thanks, Much clearer now. This will go in my notebook. $\endgroup$ – Vahram Oct 14 '15 at 21:42
  • $\begingroup$ I find this strange... When $\Delta u = 0$ then obviously $\Delta y = 0$ and hence we can write $\Delta y/\Delta x = \text{ anything }\times \Delta u/\Delta x$. Why choose specifically $dy/du$? A proper justification of $\Delta u = 0$ case is given in my answer. In this case we have $du/dx = 0$ and $dy/dx = 0$ so that chain rule holds. $\endgroup$ – Paramanand Singh Oct 15 '15 at 4:49
  • $\begingroup$ @ParamanandSingh : When $\Delta u=0$ then $\Delta y=0$ and we can write $\Delta y/\Delta x = \text{ anything }\times \Delta u/\Delta x$ because $\Delta u/\Delta x=0$, so this equality says $0=0$, which is true. The reason why $dy/du$ is chosen as the value of the piecewise factor when $\Delta u=0$ is that that is what makes the piecewise factor continuous. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 15 '15 at 18:38

The main problem is that you could have $\Delta u = 0$ for some $x$, which makes $\dfrac{\Delta y}{\Delta u}$ undefined.

  • $\begingroup$ Yes! Thank you. Obviously that disqualifies this as a proof, but for all other cases this analogy holds? I'm trying to make a notebook of math based on solely intuitive explanations I come up with or gather along the way. Wondering if I can include this or if it is too non-technical. $\endgroup$ – Vahram Oct 14 '15 at 20:35

The proof given in question is almost rigorous and correct but written in reverse (as pointed out by Wazul). Moreover this is a very standard proof of the chain rule. Contrary to what many students think this proof is not based on infinitesimals. But we do need to add some details.

Let $u = g(x)$ be differentiable and $y = f(u)$ be also differentiable. Then $y = f(g(x)) = (f \circ g)(x)$ and the Chain Rule says that $$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$$ or $$(f\circ g)'(x) = f'(u)g'(x) = f'(g(x))g'(x)$$ The proof in the question needs to be supplemented with proper definitions of $\Delta u, \Delta y$.

We have $\Delta u = g(x + \Delta x) - g(x)$ and $$\frac{du}{dx} = g'(x) = \lim_{\Delta x \to 0}\frac{g(x + \Delta x) - g(x)}{\Delta x} = \lim_{\Delta x \to 0}\frac{\Delta u}{\Delta x}$$ And similarly with $\Delta y = f(u + \Delta u) - f(u)$ we have $$\frac{dy}{du} = f'(u) = \lim_{\Delta u \to 0}\frac{f(u + \Delta u) - f(u)}{\Delta u} = \lim_{\Delta u \to 0}\frac{\Delta y}{\Delta u}$$ We then have \begin{align} \frac{dy}{dx} &= (f\circ g)'(x) = \lim_{\Delta x \to 0}\frac{(f \circ g)(x + \Delta x) - (f \circ g)(x)}{\Delta x}\notag\\ &= \lim_{\Delta x \to 0}\frac{f(g(x + \Delta x)) - f(g(x))}{\Delta x}\notag\\ &= \lim_{\Delta x \to 0}\frac{f(g(x) + \Delta u) - f(g(x))}{\Delta x}\notag\\ &= \lim_{\Delta x \to 0}\frac{f(u + \Delta u) - f(u)}{\Delta x}\notag\\ &= \lim_{\Delta x \to 0}\frac{f(u + \Delta u) - f(u)}{\Delta u}\cdot\frac{\Delta u}{\Delta x}\text{ (assume }\Delta u \neq 0)\notag\\ &= \lim_{\Delta x \to 0}\frac{f(u + \Delta u) - f(u)}{\Delta u}\cdot\frac{g(x + \Delta x) - g(x)}{\Delta x}\notag\\ &= \lim_{\Delta u \to 0}\frac{f(u + \Delta u) - f(u)}{\Delta u}\cdot\lim_{\Delta x \to 0}\frac{g(x + \Delta x) - g(x)}{\Delta x}\notag\\ &= f'(u)g'(x)\notag\\ &= \frac{dy}{du}\cdot\frac{du}{dx} \end{align} We have assumed in the above that $\Delta u \neq 0$ when $\Delta x \to 0$. Also by continuity of $u = g(x)$ (note that differentiability implies continuity) we have $\Delta u \to 0$ as $\Delta x \to 0$.

The above argument fails when $\Delta u = g(x + \Delta x) - g(x)$ vanishes for infinitely many values of $\Delta x$ as $\Delta x \to 0$. In this case we we have $du/dx = g'(x) = 0$. Why? Because if $g'(x) \neq 0$ then the ratio $\Delta u / \Delta x \neq 0$ for all small values of $\Delta x$ and hence $\Delta u \neq 0$ for all small values of $\Delta x$.

Hence if $\Delta u = 0$ for infinitely many small values of $\Delta x$ then $du/dx = g'(x) = 0$. We show that in this case $dy/dx = 0$. Clearly for those values of $\Delta x$ for which $\Delta u = 0$ we also have $\Delta y = f(u + \Delta u) - f(u) = 0$ so that the ratio $\Delta y/\Delta x = 0$. For values of $\Delta x$ where $\Delta u \neq 0$ we know that $\Delta y/\Delta u$ is bounded (because the derivative $dy/du$ exists) and the ratio $\Delta u/\Delta x$ can be made arbitrarily small (because its limit is $du/dx = 0$). Hence the overall product $$\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x}$$ can be made arbitrarily small by choosing $\Delta x$ sufficiently small. It follows that $\Delta y / \Delta x$ can be made arbitrary small for all sufficiently small values of $\Delta x$. It follows that $$\frac{dy}{dx} = \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = 0$$ and therefore the chain rule holds true in this case also.

Note that most common textbooks of calculus omit the discussion of the case when $\Delta u = 0$. Also it is much better to type and understand if $\Delta x$ is replaced by $h$ and $\Delta u$ is replace by $k$. However I have tried to stick to the notation used by OP.


This is not a proof, because you start with the formula you want to prove and imply a true statement. But a true statement can by implied by a false statement. You would have to start with

$$ \frac{dy}{dx}=\frac{dy}{dx} $$

and then prove

$$ \frac{dy}{dx}= \frac{dy}{du}\cdot\frac{du}{dx} $$

and not the other way around.

  • $\begingroup$ Isn't that exactly how all proofs we do in a College Trig class work though? Assume an identity is true, manipulate one side into looking exactly like the other, and there you have your proof. What's different about this? $\endgroup$ – Vahram Oct 14 '15 at 20:32
  • $\begingroup$ You're right of course that starting with what you want to prove and deducing something true is not a proof. But that's not the problem here - if that were the major problem then what he wrote could easily be made into a valid proof, just by rewriting it. $\endgroup$ – David C. Ullrich Oct 14 '15 at 20:33
  • $\begingroup$ I disagree, because you cannot be sure that you can seperate the two limits. $\endgroup$ – Wauzl Oct 14 '15 at 20:38

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