AP Calculus AB and BC

Chain Rule vs u-Substitution

Q: Do I have to change the limits in a definite integral?

You have two correct options. Convert the limits to u values and evaluate in u, or keep the original limits, substitute back to x, and evaluate then. Mixing the two, evaluating a u expression at the x limits, is what costs points.

They are the same rule read in opposite directions: the chain rule multiplies by the derivative of the inner function, and u-substitution divides that factor back out. In this reverse-chain-rule form, substitution works only when the inner derivative is already sitting in the integrand as a factor.

Chain rule

Use when: You are differentiating and the expression is a composite, one function evaluated inside another, with both the inner and the outer function differentiable at the point in question, so the derivative of the inner function comes along as a factor.

u-Substitution

Use when: You are integrating and, in the reverse-chain-rule form, the derivative of the inner function is already present in the integrand as a multiplying factor, up to a constant.

Side by side

	Chain rule	u-Substitution
Direction	Differentiation	Integration
Statement	$\frac{d}{dx}\,f(g(x)) = f'(g(x))\,g'(x)$	$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$
The inner derivative	Produced automatically as a factor	Consumed by $du = g'(x)\,dx$ , so in this reverse-chain-rule form it must already be present
One example, both ways	$\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$	$\int 2x\cos(x^2)\,dx = \sin(x^2) + C$
When it fails	When either piece is not differentiable, since the rule needs $g$ differentiable at $x$ and $f$ differentiable at $g(x)$ ; $\sin\lvert x\rvert$ has no derivative at $0$	Whenever $g'(x)$ is missing, as in $\int \cos(x^2)\,dx$

Both rules are statements about the same object, a composite $f(g(x))$ . Differentiating one produces the derivative of the inner function as a factor, so the chain rule hands you $g'(x)$ whether you asked for it or not. Integration reads that statement right to left, which means the $g'(x)$ has to be supplied by the integrand before substitution can cancel it against $du$ .

\frac{d}{dx}\,F(g(x)) = F'(g(x))\,g'(x) \qquad \Longleftrightarrow \qquad \int F'(g(x))\,g'(x)\,dx = F(g(x)) + C

The test takes a few seconds. Name the inner function, differentiate it, and look for that derivative in the integrand up to a constant multiple. In $\int x\,e^{x^2}\,dx$ the inner function is $x^2$ , its derivative is $2x$ , and the integrand carries an $x$ , which is $2x$ times $\frac{1}{2}$ . So $u = x^2$ with $du = 2x\,dx$ turns the problem into $\frac{1}{2}\int e^u\,du = \frac{1}{2}e^{x^2} + C$ . That test is the right one because it is the chain rule read backwards; substitutions where you solve back for $x$ and $dx$ instead, such as $u = \sqrt{x}$ in $\int e^{\sqrt{x}}\,dx$ , can still succeed without the inner derivative present, but they are no longer the chain rule reversed.

The mistake: manufacturing the missing factor

Because the chain rule creates the inner derivative for free, students expect integration to create it too and write $\int e^{x^2}\,dx = \frac{1}{2x}\int 2x\,e^{x^2}\,dx$ . Only constants move across an integral sign, never a variable, so that line is false. When the inner derivative is genuinely absent, substitution is unavailable, and $\int e^{x^2}\,dx$ has no elementary antiderivative at all.

Frequently asked questions

Do I use the chain rule or u-substitution?

Look at the operation, not the expression. Differentiating a composite calls for the chain rule every time, provided both the inner and the outer function are differentiable at the relevant points. Integrating calls for substitution in its reverse-chain-rule form only when the derivative of the inner function appears as a factor, so check for it before committing to the method.

Does u-substitution undo the chain rule?

In its reverse-chain-rule form, yes. Any substitution you finish can be verified by differentiating your answer with the chain rule and watching the $du$ factor reappear. That check takes seconds and catches a dropped constant immediately.

Do I have to change the limits in a definite integral?

You have two correct options. Convert the limits to $u$ values and evaluate in $u$ , or keep the original limits, substitute back to $x$ , and evaluate then. Mixing the two, evaluating a $u$ expression at the $x$ limits, is what costs points.

In the CED: Unit 3: Chain Rule, Implicit, and Inverses, Unit 6: Integration and Accumulation