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Integration Strategy · Tutorial 21 of 998

Differentiation As An Antiderivative Check

Use differentiation to test proposed antiderivatives and diagnose common errors from the structure of the derivative.

Beginner 10 min read

What You'll Learn

  • Define an antiderivative on an interval and explain why differentiation tests a candidate.
  • Check polynomial, exponential, logarithmic, and trigonometric candidates step by step.
  • Apply the chain rule with its hypotheses stated correctly.
  • Trace a mismatch to a sign, coefficient, factor, or inner-derivative issue.
  • Explain why antiderivatives of the same function on an interval differ by a constant.

Why Differentiate a Proposed Antiderivative?

Finding an antiderivative can involve several choices: a standard formula, an algebraic rewrite, or a substitution suggested by the expression. Once a candidate has been found, there is a direct way to test it: differentiate the candidate and compare the result with the integrand.

This check is useful because differentiation reverses the relationship that defines an antiderivative. It can confirm a correct answer, but it can also expose a missing coefficient, a sign error, or an omitted factor from the chain rule.

Key idea: A proposed function \(F\) is an antiderivative of \(f\) on an interval if differentiating \(F\) gives \(f\) at every point of that interval.

Definition and the Verification Rule

Definition: Let \(I\) be an interval and let \(f\) be a function defined on \(I\). A function \(F\) is an antiderivative of \(f\) on \(I\) if \(F\) is differentiable on \(I\) and \(F'(x)=f(x)\) for every \(x\in I\).

The definition gives the verification rule immediately. To check a candidate, differentiate it on the interval under consideration and compare its derivative with the original integrand. The two expressions must agree throughout the interval, not just at a few sample values.

Theorem: If \(F\) is differentiable on an interval \(I\), then \(F\) is an antiderivative of \(f\) on \(I\) exactly when \(F'(x)=f(x)\) for every \(x\in I\).

This theorem is the definition stated as a practical test. The requirement that the equality hold throughout the interval matters: matching at one point, or at several selected points, does not establish that two functions are equal everywhere.

Why the Constant of Integration Does Not Affect the Check

If \(F\) is an antiderivative of \(f\), then \(F(x)+C\) is also an antiderivative for any constant \(C\), because the derivative of a constant is zero:

$$ \frac{d}{dx}\bigl(F(x)+C\bigr)=F'(x)+0=f(x). $$

On an interval, this accounts for all possible differences between antiderivatives. If \(F\) and \(G\) are both antiderivatives of the same function \(f\), then \(H=F-G\) is differentiable and \(H'(x)=F'(x)-G'(x)=0\). For any two points \(a,b\in I\), the Mean Value Theorem applies to \(H\) on the closed segment between them. It gives a point \(c\) between \(a\) and \(b\) such that

$$ H(b)-H(a)=H'(c)(b-a)=0. $$

Thus \(H(b)=H(a)\) for every pair of points in \(I\), so \(H\) is constant there. Therefore \(F\) and \(G\) differ by a constant. This is why a check focuses on the derivative, not on the value of the proposed expression at a particular point.

Worked Checks

Worked Example: Checking a Polynomial Antiderivative

Check whether \(F(x)=x^3-2x^2+5x+7\) is an antiderivative of \(f(x)=3x^2-4x+5\) on \(\mathbb R\).

Differentiate each term. The derivative of \(x^3\) is \(3x^2\), the derivative of \(-2x^2\) is \(-4x\), the derivative of \(5x\) is \(5\), and the derivative of \(7\) is \(0\). Therefore,

$$ F'(x)=3x^2-4x+5+0=3x^2-4x+5. $$

This equals \(f(x)\) for every real \(x\), so \(F\) is an antiderivative of \(f\) on \(\mathbb R\). The added \(7\) has no effect on the check because its derivative is zero.

Check: Substituting the derivative expression back into the defining condition gives \(F'(x)=f(x)\), as required.

Worked Example: Checking an Exponential with a Linear Inner Function

Check whether \(F(x)=\frac{1}{2}e^{2x}\) is an antiderivative of \(f(x)=e^{2x}\) on \(\mathbb R\).

The chain rule applies here because the outer function \(E(u)=e^u\) is differentiable at every real input, and the inner function \(g(x)=2x\) is differentiable for every real \(x\). The derivative of \(e^{2x}\) is \(e^{2x}\cdot 2\). Including the constant coefficient gives

$$ F'(x)=\frac{1}{2}\left(e^{2x}\cdot 2\right) =\frac{2}{2}e^{2x} =e^{2x}. $$

Thus \(F'(x)=f(x)\) throughout \(\mathbb R\), so the proposed function passes the check.

Check: The factor \(\frac{1}{2}\) in \(F\) cancels the factor \(2\) produced by differentiating the inner function \(2x\).

Worked Example: Finding a Missing Coefficient in a Logarithmic Candidate

For \(x>0\), check the candidate \(G(x)=\ln(x^2+4)\) against the integrand \(f(x)=\frac{x}{x^2+4}\). If it fails, find a constant multiple of \(G\) that passes.

The logarithm is defined because \(x^2+4>0\) for every real \(x\). To differentiate the composite function, the outer function \(L(u)=\ln u\) must be differentiable at the inner values \(u=x^2+4\), and the inner function \(x^2+4\) must be differentiable. Both conditions hold: \(x^2+4>0\), and its derivative is \(2x\). The chain rule gives

$$ G'(x)=\frac{1}{x^2+4}\cdot 2x =\frac{2x}{x^2+4}. $$

This is twice the target integrand, not the target integrand itself. Thus \(G\) is not an antiderivative of \(f\). For a constant multiple \(A G(x)\), differentiation gives

$$ \frac{d}{dx}\left(A\ln(x^2+4)\right) =A\frac{2x}{x^2+4}. $$

To make this equal to \(\frac{x}{x^2+4}\), choose \(A\) so that \(2A=1\), which gives \(A=\frac{1}{2}\). The corrected candidate is \(F(x)=\frac{1}{2}\ln(x^2+4)\), and its derivative is

$$ F'(x) =\frac{1}{2}\cdot\frac{2x}{x^2+4} =\frac{x}{x^2+4}. $$

Check: The corrected derivative exactly equals the given integrand for every \(x>0\) (and in fact wherever the displayed expressions are defined).

Worked Example: Diagnosing a Sign Error

Check whether \(F(x)=\cos x\) is an antiderivative of \(f(x)=\sin x\) on \(\mathbb R\). If it is not, correct the sign.

The derivative of cosine is negative sine, so

$$ F'(x)=\frac{d}{dx}(\cos x)=-\sin x. $$

The result has the opposite sign from the target \(\sin x\). Therefore \(\cos x\) fails the check. Changing its sign gives \(G(x)=-\cos x\), and

$$ G'(x)=\frac{d}{dx}(-\cos x) =-(-\sin x) =\sin x. $$

Check: The corrected candidate differentiates to the original integrand, so \(-\cos x\) is an antiderivative of \(\sin x\) on \(\mathbb R\).

Reading the Derivative to Locate an Error

A failed check is informative when the derivative is compared with the integrand symbolically. Look at which part of the expression differs rather than treating the result as simply “wrong.”

What the derivative shows Likely issue to inspect
The signs are opposite A negative sign may have been lost or introduced.
The expression has the right form but is multiplied by a constant A coefficient may be missing or incorrect.
An extra factor appears alongside a composite expression Check the derivative of the inner function in the chain rule.
A factor or term is absent Differentiate the proposed expression term by term and compare each part.

These are diagnostic clues, not substitutes for differentiation. A coefficient mismatch can sometimes be repaired by adjusting a constant multiple, as in the logarithmic example. A more complicated mismatch may require revisiting how the candidate was obtained. In either case, the proposed function is verified only after its derivative agrees with the integrand on the interval.

The Chain Rule Must Have Its Hypotheses

For composite expressions, the check often depends on the chain rule. At a point \(x_0\), if \(g\) is differentiable at \(x_0\) and \(F\) is differentiable at \(g(x_0)\), then the composite function is differentiable at \(x_0\) and

$$ \frac{d}{dx}F(g(x))\bigg|_{x=x_0} =F'(g(x_0))g'(x_0). $$

Both differentiability conditions matter. Differentiability of \(g\) alone does not guarantee that \(F(g(x))\) is differentiable. For instance, \(g(x)=x\) is differentiable at \(0\), but \(F(u)=|u|\) is not differentiable at \(g(0)=0\); accordingly, \(F(g(x))=|x|\) is not differentiable at \(0\). When checking a candidate on an interval, verify that the outer function is differentiable at the relevant inner values and that the inner function is differentiable there.

A Common Mistake: Checking Only the Familiar Part

A frequent error is to recognize the outer form of a candidate and forget that differentiating a composite expression also differentiates its inner function. For example, the derivative of \(e^{2x}\) is not merely \(e^{2x}\). The inner function \(2x\) contributes its derivative \(2\):

$$ \frac{d}{dx}e^{2x}=e^{2x}\cdot 2=2e^{2x}. $$

A wrong check would omit that factor and claim \(\frac{d}{dx}e^{2x}=e^{2x}\). The correction is to identify the inner function \(g(x)=2x\), compute \(g'(x)=2\), and include it in the chain rule. This is exactly why differentiating a proposed antiderivative is more reliable than relying on a remembered visual pattern.

Why This Check Matters

Integration methods can produce expressions that look plausible but contain a small algebraic error. Differentiation gives a local test of the entire result: constants disappear, signs become visible, coefficients can be compared, and derivatives of inner functions supply factors that might otherwise be overlooked.

The check is also economical. It does not require repeating the method used to find the candidate. Once a proposed antiderivative is available, differentiate it and compare. If the expressions match throughout the relevant interval, the definition confirms the result; if not, the mismatch points toward what needs attention.

Check Your Understanding

For each question, use the derivative as the test and state the interval on which your conclusion applies.

  1. Is \(x^4-3x^2+8\) an antiderivative of \(4x^3-6x\) on \(\mathbb R\)? Show the derivative.
  2. Differentiate \(\frac{1}{3}e^{3x}\). What integrand does it provide?
  3. Does \(\ln(x^2+1)\) have derivative \(\frac{x}{x^2+1}\)? If not, what constant multiple of the logarithm gives that derivative?
  4. Check whether \(\sin x\) is an antiderivative of \(\cos x\). If it fails, give a corrected candidate.
  5. State both differentiability conditions needed to apply the chain rule to \(F(g(x))\) at a point \(x_0\).

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