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I understand that the derivative of a function can be evaluated from first principles. However I do not understand why dy/dt x dt/dx = dy/dx. Surely these terms cannot be treated in the same way in which you would multiply fractions. Could somebody please explain this intuitively?

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  • $\begingroup$ You are right that they are not fractions. However, they behave like fractions so much that Leibniz thought they were fractions, and that is reflected in his notation. Back to the question (at least the first principles part): have you tried just applying the definition and see what you get? $\endgroup$ – Arthur Apr 11 '17 at 15:49
  • $\begingroup$ I have tried, but did not get very far $\endgroup$ – Nav Hari Apr 11 '17 at 16:06
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To explain the chain rule intuitively, we first need an intuitive notion of derivative, and to aid that, I will introduce an intuitive notion of function.

In this answer, a function is a mechanism that takes input by some manner of indication (I like to think about pointing) on one number line, and gives output by indicating (I like to think about a red dot appearing) on a second number line. As one moves the input, the output moves correspondingly in real time.

A derivative at a point $p$ in this model can be thought about the following way: first point at $p$ on the first number line, and note where $f(p)$ appears on the other number line. Then point at somewhere relatively close to $p$, and note how much the function moves its point on the second number line. As you choose your second point closer and closer to $p$, $f$ will indicate points closer and closer to $f(p)$ (this is what continuity means). These two distances are approximately proportional (this is the definition of differentiable), and they will get closer and closer to being proportional the closer to $p$ you move about. The proportionality constant that this almost-proportional relationship tends towards, is the derivative of $f$ at $p$, which we write $f'(p)$.

In other words, $f'(p)$ is the real number so that for any small number $q$, if we point at $p+q$, then the function indicates a point very close to $f(p)+f'(p)\cdot q$. ("Very close" does have a precise definition, but I'm leaving it out for now.)

Now let's look at the chain rule. Say we have the function $h(x)=f(g(x))$, and we're interested in the derivative of $h$ at $x=p$, expressed using $f$ and $g$.

First, how does the chaining work in our interpretation of functions? First, we indicate our input on one number line, and then $g$ indicates its output on a second number line. Then, $f$ uses that indication to produce its own output on a third number line. This entire mechanism is how $h$ takes input on the first number line and produces its output on the third line.

Finally, we come to the chain derivative. Let's start at a point $p$. On the intermediate second line, $g$ indicates the point $g(p)$, and on the final, third number line, $f$ (and therefore $h$) indicates $f(g(p))$. Then, we point to $p+q$ for some small $q$. By definition of derivative, on the second number line, $g$ indicates a number very close to $g(p)+g'(p)\cdot q$. Since $g'(p)\cdot q$ is still a small number, this means that $f$, again by our definition of derivative, indicates a point very close to $f(g(p))+f'(g(p))\cdot g'(p)\cdot q$.

Rereading the last point in terms of $h$, that means that if we give it $p+q$ as input, then $h$ gives us something very close to $f(g(p))+f'(g(p))g'(p)q$. But by definition of the derivative, it is supposed to give us something very close to $h(p)+h'(p)q$. We easily see that $f(g(p))$ corresponds to $h(p)$, add $q$ is still $q$. That means that $f'(g(p))g'(p)$ must correspond to $h'(t)$.

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  • $\begingroup$ This is very nice. +1 $\endgroup$ – Andres Mejia Apr 11 '17 at 17:04
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I do not think I can do nearly as well as Arthur's answer in the way of intuition. However, one can prove it from "first principles:"

$$\lim_{x \to a}\frac{f(g(x))-f(g(a))}{x-a}=\lim_{x \to a} \frac{f(g(x))-f(g(a))}{g(x)-g(a)} \cdot \frac{g(x)-g(a)}{x-a}$$ where the second expression becomes $f^{\prime}(g(a)) \cdot g^{\prime}(a)$ since limits "distribute" over multiplication (since it is continuous), and then applying definitions.

My way of thinking about this is via "co-ordinates." If we "normalize" the $x$-axis to really be a function, then the chain rule says "the first is how fast something is moving with respect to this new function" and then you multiply by "how fast" (or sloped) our function was with respect to the stand-still $x$-axis.

Another less geometric way is to think of the graph $(x,f(x))$ of a single function and then define a function on this graph to merely be an output, a number associated to $g:f(x) \mapsto g(f(x)) $. With respect to $f(x)$, the derivative is just $g^{\prime}f(x)$, but relative to the axes, you also want to see how "sloped" or "fast" $f(x)$ is growing and multiply said rates.

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