Rates multiply along a composition. If the inner function changes three times as fast as its input and the outer one changes twice as fast as its input, the composition changes six times as fast — and the outer rate must be measured where the inner function actually lands, not where the input is. That is the whole rule, and the only difficulty in proving it is that the inner step may be zero.

Gears in a train. The final shaft turns at the rate of the last gear times the rate of the one before, and the ratio of each gear is read at the tooth actually engaged. Multiplying the ratios is the chain rule; reading the last ratio at the wrong tooth is the mistake it warns against.

If $f$ is differentiable at $c$ and $g$ is differentiable at $f(c)$, then $g\circ f$ is differentiable at $c$ and $(g\circ f)'(c)=g'\big(f(c)\big)\,f'(c)$. The outer derivative is evaluated at the image of the point. The naive proof multiplies and divides by the inner step, which may be zero; the repair is to use a function that equals the outer difference quotient where that is defined and the outer derivative where it is not.

• The outer derivative is taken at $f(c)$, never at $c$. Writing $g'(c)$ is the standard mistake and gives a wrong answer whenever $f(c)\ne c$.

Define a helper function on the values of the inner function: it equals the outer difference quotient wherever the step is not zero, and the outer derivative where it is. That helper is continuous at the image of the point, precisely because the outer derivative is the limit of the quotient there. With the helper in hand the identity holds for every input including the awkward ones, and the limit is then the product of two limits, one of which uses the continuity of the inner function.

Define a helper that patches the outer quotient at the one point where it is undefined.

Because the outer derivative is exactly the limit of that quotient, the patched value is the limit.

This identity holds for every input, including those where the inner step is zero.

Divide by the outer step, which is not zero because the limit avoids the point.

The inner function is continuous, so the helper is evaluated in the limit at the image of the point.

The outer derivative is taken where the inner function lands.

• Rates multiply along a composition.
• The outer rate is read at the image of the point.
• The naive proof fails because the inner step may vanish arbitrarily near the point.
• A helper function patching the outer quotient repairs it.

Great work! You now know how to:

• Apply the chain rule and read each rate at the right point
• Handle a composition of three functions
• Say why the naive proof fails and what repairs it
• Avoid evaluating the outer derivative at the wrong place