The chain rule AB & BC

Write $y = \sin(3x)$ as two steps. First the inside does its job, $u = 3x$. Then the outside acts on whatever came out, $y = \sin u$. Nothing about that is unusual; what matters is what happens to rates when you stack the steps.

Move $x$ by a small amount. Because $u = 3x$, the middle quantity moves three times as fast. The sine machine does not know or care that its input is racing; it just responds at its own rate, $\cos u$, per unit of $u$. So the output rate is the outer rate multiplied by how fast the inside is feeding it:

$$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = \cos(3x)\cdot 3.$$

That second factor is the entire content of the chain rule, and dropping it is the single most common error in differential calculus. It is invisible in the one case where the inside is just $x$, because then $du/dx = 1$ and the missing factor happens to be harmless. Every other time, it is not.

For $f(g(x))$ the rule reads $\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$. Two details in that expression carry all the weight. The outer derivative is evaluated at the inside function, not at $x$: it is $f'(g(x))$, never $f'(x)$. And the inner derivative multiplies on, untouched.

A reliable procedure:

1. Ask what you would do last if you were evaluating the expression at a number. That is the outer function.
2. Differentiate the outer function, leaving its input exactly as it stands.
3. Multiply by the derivative of that input.
4. If the input is itself a composition, repeat from step 1 on it.

So $\frac{d}{dx}(x^2+1)^5 = 5(x^2+1)^4\cdot 2x$. To evaluate at a number you would square, add one, then raise to the fifth, so the fifth power is outermost. Note the inside stays $x^2+1$ inside the fourth power. Writing $5(2x)^4$ instead means differentiating the inside and then feeding that to the outer rule, which is not what the rule says.

Nothing changes when the nesting goes deeper, but the bookkeeping gets easier to abandon halfway. For $y = \sin^2(3x)$ there are three machines: square, then sine, then triple. Peeling outward to inward gives

$$\frac{dy}{dx} = \underbrace{2\sin(3x)}_{\text{square}}\cdot\underbrace{\cos(3x)}_{\text{sine}}\cdot\underbrace{3}_{\text{triple}} = 6\sin(3x)\cos(3x).$$

The characteristic failure is stopping one layer short and writing $2\sin(3x)\cos(3x)$. Everything visible in that answer is correct; the missing piece is the factor from the layer nobody got to. A quick audit: count the machines, then count the factors. They must match.

The rules are not interchangeable, and choosing between them is a reading problem, not a calculus problem. Read the structure before reaching for a rule:

1. $x^2\sin x$ is a product: two independent functions multiplied. Product rule.
2. $\sin(x^2)$ is a composition: one function's output is the other's input. Chain rule.
3. $(x+1)^3$ is a composition, not a product. The exponent is not a second factor.
4. $x^2\sin(3x)$ is a product whose second factor needs the chain rule. Both rules, in that order.

The tell is parentheses doing real work. In $\sin(x^2)$ the $x^2$ sits inside the sine and cannot be pulled out front. In $x^2\sin x$ the two pieces are simply multiplied and could be written in either order. When both structures appear at once, apply the product rule at the top level and let the chain rule handle whichever factor needs it.