## The Intuitive Notion of the Chain Rule |
HELP |

In the applet we see an `x`-wheel, a `u`-wheel, and a `y`-wheel. You can change the
speed of the `x`-wheel, and you
can connect the wheels with belts and change their radii.
We'll use this model to explore the chain rule and try to get an intuitive
understanding of where the formula comes from.

Suppose that `u` is a function of `x`. So, as `x` changes, it causes `u` to change. We might
ask ourselves, *What is the rate of change of `u`, relative to the rate of change of `x`?*
Does `u` change twice as much as `x`? If you increase `x` a little bit does `u` increase ten
times that amount? Half that amount? Maybe as `x` increases, `u` decreases - in this
case the rate of change of `u` is negative, relative to the rate of change of `x`. In any event,
here's the notation:

If `u` is a function of `x`, then `du/dx` denotes the rate of change of `u`
*relative to* the rate of change of `x`. Informally, `u` changes `du/dx`
times as fast as `x` does.

- If necessary, reload this web page to reset the applet to its initial configuration. Click the checkbox to connect the `x`-wheel and the `u`-wheel. How is `du/dx` related to the motion of the wheels?
- Vary the speed of the `x`-wheel with the top slider. How is `du/dx` affected? Explain.
- Disconnect the belt between the `x`-wheel and `u`-wheel. Explain why `du/dx = 0`.
- Connect the three wheels. Now `dy/du` is the speed of the `y`-wheel relative to the speed of the `u` wheel, and `du/dx` is the speed of the `u`-wheel relative to the speed of the `x` wheel. How can we figure out the speed of the `y`-wheel relative to the speed of the `x`-wheel, that is, `dy/dx` ? Examine the situation, make your best guess, then click the "Reveal dy/dx" checkbox to check your intuition.
- The values of `x`, `u`, and `y` are recorded at the bottom of the applet. How do the the values of the variables affect the values of `du/dx`, `dy/du`, and `dy/dx` ? Explain.
- Play with various wheel sizes and connections. Is it always the case that `dy/dx = dy/du du/dx` ? Explain.

If `y` is a quantity that depends on `u`, and `u` is a
quantity that depends on `x`, then ultimately, `y` depends on `x` and
`dy/dx = dy/du du/dx`.

**Example**. Suppose `y = u^10` and `u = x^4 + x`. Then [`dy/dx = dy/du du/dx = 10u^9 (4x^3 + 1) = 10(x^4 + x)^9 (4x^3 + 1)`] Notice how in the last equality we write out `u` in terms of `x` so that `dy/dx` is expressed entirely in terms of `x`.**Example**. If `f(x) = (x^4 + x)^10` how can we compute `df/dx` ?

Let `u = x^4 + x` and let `y = f(x)`. Then `y = u^10` and we can solve the problem just like in the previous example. Whenever we have a composition of functions, we typically let `u` represent the "inside" function.**Example**. If `f(x) = g(h(x))` how can we compute `f'(x)` ?

Let `u = h(x)` and let `y = f(x)`. (See how `u` represents the "inside" function.) Then `y = g(u)`. Now [`f'(x) = dy/dx = dy/du du/dx = g'(u) h'(x) = g'(h(x)) h'(x)`]

If `f(x) = g(h(x))`, then [`f'(x) = g'(h(x)) h'(x)`]

Alternatively,
[`d/dx f(g(x)) = f'(g(x)) g'(x)`]