AP Calculus AB and BC

Chain Rule Step Trainer: You Pick the Decomposition

Q: Do I need to rewrite roots and reciprocals before decomposing?

Yes, and it makes the outer function obvious. Rewrite \sqrt{u} as u^{1/2} and \dfrac{1}{u} as u^{-1} so the last operation reads as a power. Then g(u) = u^{1/2} or g(u) = u^{-1} is the outer function and the power rule handles g', with the chain rule multiplying by the inner derivative.

The chain rule's hard part is decomposition: naming the outer and inner function before you differentiate. Use the last-operation test: whatever you compute last is the outer function, everything inside it is the inner. Differentiate the outer with the inside untouched, then multiply by the inner's derivative.

Problem

1 / 10

Differentiate

y = \sin(3x^2)

Step 1: which is the outer function? (Take the inner function to be everything inside the last operation.)

The chain rule itself is one line. The step that actually decides whether you get the derivative right is decomposition: splitting a composite function like $f(x) = g(h(x))$ into an outer function $g$ and an inner function $h$ before you differentiate anything. This trainer makes you commit to that split from a set of choices, then assembles the derivative one step at a time so you can see how your pick drives every factor. That is the exact skill AP Calculus Topic 3.1 is built around: skill 1.C, choosing a procedure based on how an expression is put together rather than memorizing a formula.

f'(x) = g'(h(x)) \cdot h'(x)

Read the formula as a recipe with three slots: the derivative of the outer function evaluated at the untouched inside, then multiplied by the derivative of the inside. Every slot depends on naming $g$ and $h$ correctly first, which is why the trainer asks you to choose the outer and inner function before it will build anything. To find them, use the last-operation test: imagine plugging in a number and running the arithmetic. Whatever you would do last is the outer function; everything you did before that is the inner function.

The last-operation test

For $\sqrt{7x - 1}$ you would compute $7x - 1$ first and take the square root last, so the outer function is the square root and the inner is $7x - 1$ . For $e^{x^2}$ you would compute $x^2$ first and exponentiate last, so the outer function is $e^{u}$ and the inner is $x^2$ . Say the split out loud before you differentiate.

Once the split is named, the single habit that keeps chain-rule work clean is leaving the inside untouched while you differentiate the outside. The derivative of $(3x^2 + 5x)^4$ keeps $3x^2 + 5x$ intact as $4(3x^2 + 5x)^3$ ; only afterward do you multiply by the inner derivative $6x + 5$ . The trainer enforces this order, so a wrong pick shows up immediately as a factor that does not belong, which is the fastest way to feel the difference between the two candidate functions.

Expression	Outer function (last step)	Inner function
$(3x^2 + 5x)^4$	raise to the 4th power, $u^4$	$3x^2 + 5x$
$\sin(4x)$	sine, $\sin u$	$4x$
$e^{x^2}$	exponential, $e^{u}$	$x^2$
$\sqrt{7x - 1}$	square root, $u^{1/2}$	$7x - 1$
$\sin^2 x$	square, $u^2$	$\sin x$

The last row is the classic trap: $\sin^2 x$ means $(\sin x)^2$ , so squaring is the outer function and $\sin x$ is the inner, which is the opposite decomposition from $\sin(x^2)$ . Working several of these in the trainer trains your eye to read the notation before you commit. Aim to name the outer and inner function correctly on sight; the assembly that follows is mechanical once the split is right, and getting it solid here pays off in implicit differentiation (Topic 3.2), related rates (Topic 4.4), and $u$ -substitution, all of which are the chain rule wearing a different hat.

Worked example

Decompose, then assemble a square-root composition

Differentiate $f(x) = \sqrt{x^2 + 1}$ .

Rewrite the root as a power so the outer function is easy to name: $\sqrt{x^2 + 1} = (x^2 + 1)^{1/2}$ . This is the same rewrite the trainer prompts before you choose, because a radical hides the fact that the last operation is a power.
Run the last-operation test. Evaluating at a number, you would compute $x^2 + 1$ first and raise it to the $1/2$ power last. So the outer function is $g(u) = u^{1/2}$ and the inner function is $h(x) = x^2 + 1$ , giving $f(x) = g(h(x))$ .
Differentiate the outer function, leaving the inside untouched. Since $g(u) = u^{1/2}$ , we have $g'(u) = \tfrac{1}{2} u^{-1/2}$ , so $g'(h(x)) = \tfrac{1}{2}(x^2 + 1)^{-1/2}$ . Notice the inside stays exactly $x^2 + 1$ .
Differentiate the inner function on its own: $h'(x) = 2x$ .
Multiply the two pieces per the chain rule: $f'(x) = g'(h(x)) \cdot h'(x) = \tfrac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x$ .
Simplify. The $\tfrac{1}{2}$ and the $2x$ combine to $x$ , and the negative exponent moves the factor into a denominator: $f'(x) = \dfrac{x}{\sqrt{x^2 + 1}}$ .

$f'(x) = \dfrac{x}{\sqrt{x^2 + 1}}$

Frequently asked questions

How do I decide which function is the outer and which is the inner?

Use the last-operation test. Imagine evaluating the expression at a number and watch the order of operations: whatever you would do last is the outer function, and everything you computed before that is the inner function. For $\cos(3x - 1)$ you subtract inside first and take cosine last, so cosine is outer and $3x - 1$ is inner.

What happens if I pick the outer and inner functions backwards?

You differentiate the wrong function and the factors come out mismatched. If you treated $\sin(x^2)$ as though squaring were the outer step (as in $\sin^2 x$ ), you would differentiate $(\sin x)^2$ and get $2\sin x \cos x$ , instead of the correct derivative $2x\cos(x^2)$ , which is a different function entirely. The trainer surfaces this right away so you can see why the split has to come first.

Do I need to rewrite roots and reciprocals before decomposing?

Yes, and it makes the outer function obvious. Rewrite $\sqrt{u}$ as $u^{1/2}$ and $\dfrac{1}{u}$ as $u^{-1}$ so the last operation reads as a power. Then $g(u) = u^{1/2}$ or $g(u) = u^{-1}$ is the outer function and the power rule handles $g'$ , with the chain rule multiplying by the inner derivative.