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I've been wondering if is there an easy way to explain derivative's Chain Rule, since it's such a fundamental topic in Calculus and people struggle to understand the first time that they get in touch with the subject (like I did).

How would you explain it to your grandmother?

Thank you!

bru1987
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    $$\dfrac{\mathrm dy }{\mathrm dx } = \dfrac{\mathrm dy }{\mathrm dz } \times \dfrac{\mathrm dz}{\mathrm dx }$$ Grandma, see how the $\mathrm dz$'s cancel out. – Ishfaaq Aug 13 '15 at 12:08
  • Hi @Ishfaaq can you produce a simple verbal explanation, without formulas? – bru1987 Aug 13 '15 at 12:09
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    Related questions include Chain Rule Intuition. Do the answers there help? – Andrew D. Hwang Aug 13 '15 at 12:11
  • The Usain Bolt analogy used there is really intuitive, thank you @AndrewD.Hwang – bru1987 Aug 13 '15 at 12:13
  • Can't quite understand the downvote. This is a legitimate question related to teaching. @bru1987 the answer there by Mariano about linear approximations is also very enlightening. – Ishfaaq Aug 13 '15 at 12:17
  • Trying to describe it without formulae strikes me as silly project. In any case, you might a more receptive audience on http://matheducators.stackexchange.com/ – Simon S Aug 13 '15 at 12:20
  • That's true @Ishfaaq, even though I think the Usain bolt analogy resonates better with teen students, it is a good explanation, intuitive also to explain. Thank you. – bru1987 Aug 13 '15 at 12:21
  • I'm not trying not to use formulae @SimonS, just use it as an introduction before the formulae begins. For example, the first sentences I could say to introduce the topic. – bru1987 Aug 13 '15 at 12:23

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My Grandmother was a Professor of Functional Analysis so I wouldn't even dare. However, if I did, I'd probably say if $f$ and $g$ are functions then the chain rule tells you how to write down the derivative of their composition.

That is; to evaluate $(f\circ g)'$ defined by $$(f\circ g)'=(f'\circ g)\cdot g'$$

Autolatry
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If you are asking for the purpose of learning, not the proof, then take the example of $\frac{d}{dx}\sin(\sin(x))$. Take $\sin x=u$. Now, differentiate $\sin u$, which is $\cos u$. Now remember, you took a value of $u$, now, differentiate that value of $u$, which is again, $\sin x$, which comes out to be $\cos x$/ So multiply both, $$\cos(u)\dot\sin(x)$$
$u=\sin(x)$ $$\cos(\sin(x))\sin(x)$$.

Aditya Agarwal
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  • Hi @Aditya, thank you. I was thinking a verbal explanation, something that doesn't require formulas. – bru1987 Aug 13 '15 at 12:12
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Imagine you have a few gears put together in some mechanism. You have a knob on the first and the last one, so you can turn those individually. If you think of a gear as a function, what a gear does is it takes 'input' from the gear before it, and rotates according to it's own size. Something like $$\text{twist of the (n-1)th gear }\stackrel{\text{nth gear}}\mapsto \text{twist of the nth gear }.$$

So, let's say I twist the first gear by some angle $\mathrm d \theta_1$. How is this related to the angle the last gear will twist? Something like this:

$$\mathrm d\theta_n = \prod_i^n \text{(individual change of the i-th gear)}\times \mathrm d \theta_1$$

And this is the chain rule, on an example that works because the rate of change of a gear is just a ratio of the previous to this one. Which, again, encodes the fact that you can 'cancel out' the infinitesimals to get $$\frac{df}{dx}=\frac{d f}{dy}\frac{dy}{dx}$$ in a dumb but intuitive way.

krvolok
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  • Thank you for the sophisticated explanation @krvolok. – bru1987 Aug 13 '15 at 12:31
  • It's silly but it's a nice way to explain to yourself how come you have to multiply the derivatives of each function, where each derivative is with respect to the argument of the function! – krvolok Aug 13 '15 at 12:37