Reverse Chain Rule and Substitution

Why Substitution Reverses the Chain Rule

When differentiating a composite function, the chain rule combines an outer derivative with the derivative of the inner function. Antidifferentiation can reverse that process when an integrand contains the same two pieces multiplied together.

For example, differentiating $\sin(x^2)$ produces both $\cos(x^2)$, from differentiating the outer sine function, and $2x$, from differentiating the inner expression $x^2$. If an integrand contains those factors together, the chain rule points toward an antiderivative.

Central idea: If an integrand has the form $q(g(x))g'(x)$, and $Q$ is an antiderivative of $q$, then $Q(g(x))$ is an antiderivative of the integrand. Substitution is a systematic way to express this reverse-chain-rule pattern.

The Reverse Chain Rule

Let $g$ be differentiable on an interval $I$. Suppose $q$ is defined on an interval containing the values of $g(x)$ for $x\in I$, and let $Q$ satisfy $Q'(u)=q(u)$ there. The chain rule gives

\frac{d}{dx}Q(g(x)) = Q'(g(x))g'(x) = q(g(x))g'(x).

Thus the product $q(g(x))g'(x)$ has the antiderivative $Q(g(x))$. Since indefinite integrals represent families of antiderivatives, the result includes an arbitrary constant.

Theorem: On an interval where the stated functions are defined and differentiable, if $Q'=q$, then $$ \int q(g(x))g'(x)\,dx=Q(g(x))+C. $$ Here $C$ is an arbitrary constant. The formula applies when the integrand contains the composite factor $q(g(x))$ multiplied by the derivative $g'(x)$ of its inner function.

This formula does not require that $g$ have an inverse. It uses only the chain rule and the existence of an antiderivative $Q$ for the outer factor $q$ on the relevant range.

Writing the Rule as a Substitution

The same idea is often written using a temporary variable $u$. Set $u=g(x)$. Then the differential notation $du=g'(x)\,dx$ records the factor that accompanies the composite expression. The integral becomes an integral in $u$, which can be evaluated using a known antiderivative.

u=g(x),\qquad du=g'(x)\,dx

The substitution is not a separate trick detached from differentiation: it is the reverse of the chain rule. The variable $u$ names the inner expression, and $du$ accounts for its derivative. After finding an antiderivative in $u$, replace $u$ with $g(x)$ to express the result in the original variable.

Recognize the structure.
Express the integrand as an outer function evaluated at an inner function, multiplied by the derivative of that inner function.

Rename the inner function.
Write $u=g(x)$ and $du=g'(x)\,dx$.

Integrate in $u$.
Find an antiderivative of the resulting function of $u$.

Return to $x$ and check.
Replace $u$ by $g(x)$, include $C$, and differentiate the result to verify it.

Worked Examples

Worked Example: A Power of a Composite Function

Evaluate $\int 6x(3x^2+2)^4\,dx$. The inner expression is $3x^2+2$, and its derivative is $6x$, which is the other factor in the integrand.

Set $u=3x^2+2$. Differentiating gives $du=6x\,dx$. Therefore,

\begin{aligned} \int 6x(3x^2+2)^4\,dx &=\int u^4\,du\\ &=\frac{u^5}{5}+C\\ &=\frac{(3x^2+2)^5}{5}+C. \end{aligned}

To check, differentiate the proposed antiderivative:

\begin{aligned} \frac{d}{dx}\left(\frac{(3x^2+2)^5}{5}+C\right) &=\frac{1}{5}\cdot 5(3x^2+2)^4\cdot 6x\\ &=6x(3x^2+2)^4. \end{aligned}

This is the original integrand, so the antiderivative is verified.

Worked Example: A Trigonometric Function of a Composite Expression

Evaluate $\int (2x+1)\cos(x^2+x)\,dx$. The inner expression is $x^2+x$, whose derivative is $2x+1$.

Set $u=x^2+x$. Then $du=(2x+1)\,dx$, so

\begin{aligned} \int (2x+1)\cos(x^2+x)\,dx &=\int \cos(u)\,du\\ &=\sin(u)+C\\ &=\sin(x^2+x)+C. \end{aligned}

Check by differentiating:

\begin{aligned} \frac{d}{dx}\bigl(\sin(x^2+x)+C\bigr) &=\cos(x^2+x)\frac{d}{dx}(x^2+x)\\ &=(2x+1)\cos(x^2+x). \end{aligned}

The derivative matches the integrand exactly.

Worked Example: A Reciprocal Composite Expression

Evaluate $\int \frac{2x}{x^2+1}\,dx$. The inner expression $x^2+1$ has derivative $2x$. Also, $x^2+1>0$ for every real $x$, so the logarithmic antiderivative is defined throughout the domain.

Set $u=x^2+1$, which gives $du=2x\,dx$. Then

\begin{aligned} \int \frac{2x}{x^2+1}\,dx &=\int \frac{1}{u}\,du\\ &=\ln|u|+C\\ &=\ln|x^2+1|+C\\ &=\ln(x^2+1)+C. \end{aligned}

The final equality holds because $x^2+1$ is positive. Verify by differentiation:

\begin{aligned} \frac{d}{dx}\ln(x^2+1) &=\frac{1}{x^2+1}\cdot 2x\\ &=\frac{2x}{x^2+1}. \end{aligned}

This is the given integrand, confirming the result.

Why the Method Works

The substitution formula follows directly from a complete chain-rule argument. Begin with $Q'=q$ and $u=g(x)$. Differentiating the composite function $Q(g(x))$ gives

\frac{d}{dx}Q(g(x)) = Q'(g(x))g'(x) = q(g(x))g'(x).

By definition, any function whose derivative is $q(g(x))g'(x)$ is an antiderivative of that integrand. Therefore $Q(g(x))$ is an antiderivative, and the full family of antiderivatives on the interval is $Q(g(x))+C$. This reasoning also explains why checking a proposed answer by differentiation is decisive: it tests the defining property of an antiderivative.

Recognizing the Matching Factors

A useful way to inspect an integrand is to name its roles. The inner function is the expression $g(x)$ inside another function. The outer factor is $q(g(x))$, a function applied to that inner expression. The remaining factor must be $g'(x)$ for the direct reverse-chain-rule formula to apply.

Role	What to look for
Inner function	An expression $g(x)$ appearing inside a power, trigonometric function, exponential, or other function
Outer factor	A function of the full inner expression, such as $q(g(x))$
Inner derivative	The factor $g'(x)$ multiplying the outer factor

The match must be exact for the stated formula. For instance, if the derivative of the inner expression is $2x+1$, the direct pattern contains the factor $2x+1$. The composite expression alone does not guarantee that the integral can be evaluated by this particular reverse-chain-rule step.

A Common Mistake: Forgetting the Inner Derivative

A frequent error is to integrate the outer function while ignoring the derivative introduced by the chain rule. For example, the calculation

\int \cos(x^2+x)\,dx \overset{\text{wrong}}{=} \sin(x^2+x)+C

is not justified by the reverse chain rule. Differentiate the proposed answer:

\frac{d}{dx}\sin(x^2+x) = (2x+1)\cos(x^2+x),

which contains the extra factor $2x+1$ and is not the original integrand. The matching form would be $(2x+1)\cos(x^2+x)$, as in the earlier example. The correction is not to discard the inner derivative; it is to check whether the integrand actually contains it. If it does not, this direct substitution pattern does not provide the proposed answer.

Why This Connection Matters

The chain rule is one of the basic ways derivatives of composite functions are formed. Its reverse gives a broad method for finding antiderivatives that would not be visible from a simple power rule or a direct table of basic antiderivatives. Recognizing the relationship is more important than memorizing a list of isolated formulas: the same reasoning applies whenever an outer function and its inner derivative appear together.

The most reliable final step is differentiation. A correct antiderivative must reproduce every factor in the original integrand when differentiated. This check can reveal a missing inner derivative, an incorrect outer antiderivative, or an omitted constant. It also reinforces the central point: substitution is valid because the chain rule verifies the result.

Check Your Understanding

For each question, focus on the relationship between the composite expression and the derivative of its inner function.

State the reverse-chain-rule formula using $q$, $g$, and an antiderivative $Q$ of $q$.
In $\int 4x\sin(2x^2+3)\,dx$, identify the inner function and its derivative.
Use substitution to evaluate $\int 5(5x+2)^3\,dx$, and verify the result by differentiation.
Why does $\int \cos(x^2)\,dx=\sin(x^2)+C$ fail as a reverse-chain-rule calculation?
What condition on $Q$ is needed before using $Q(g(x))$ as an antiderivative?

Reverse Chain Rule and Substitution

What You'll Learn

Why Substitution Reverses the Chain Rule

The Reverse Chain Rule

Writing the Rule as a Substitution

Worked Examples

Worked Example: A Power of a Composite Function

Worked Example: A Trigonometric Function of a Composite Expression

Worked Example: A Reciprocal Composite Expression

Why the Method Works

Recognizing the Matching Factors

A Common Mistake: Forgetting the Inner Derivative

Why This Connection Matters

Check Your Understanding

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Vol. 2: Solutions Manual

Role	What to look for
Inner function	An expression \(g(x)\) appearing inside a power, trigonometric function, exponential, or other function
Outer factor	A function of the full inner expression, such as \(q(g(x))\)
Inner derivative	The factor \(g'(x)\) multiplying the outer factor

Reverse Chain Rule and Substitution

What You'll Learn

Why Substitution Reverses the Chain Rule

The Reverse Chain Rule

Writing the Rule as a Substitution

Worked Examples

Worked Example: A Power of a Composite Function

Worked Example: A Trigonometric Function of a Composite Expression

Worked Example: A Reciprocal Composite Expression

Why the Method Works

Recognizing the Matching Factors

A Common Mistake: Forgetting the Inner Derivative

Why This Connection Matters

Check Your Understanding

Test yourself: AP Calculus practice exams

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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