Why Differentiate a Candidate Antiderivative?
An integration result is easy to misread or mistype: a constant factor may be missing, a sign may have changed, or a chain-rule factor may have been overlooked. A direct way to check a proposed antiderivative is to differentiate it and compare the result with the integrand.
This check is especially useful when a proposed expression comes from substitution. The substitution may make the integration look simple, but when the result is written back in terms of \(x\), the derivative of the inner function must still be accounted for.
The Verification Process
A good check makes the comparison explicit rather than relying on a quick glance. First identify the candidate \(F(x)\) and the target integrand \(f(x)\). Then differentiate \(F\), simplify the result, and check whether it agrees with \(f\) throughout the relevant interval.
Write down the integrand \(f(x)\) that the candidate is supposed to differentiate to.
Use the product, quotient, or chain rule whenever the expression requires it.
Rewrite the derivative in a form that can be compared directly with \(f(x)\).
If the expressions agree on the interval, the candidate is verified there. If they do not, identify the discrepancy.
The interval matters because a formula may be defined or differentiable only on part of the real line. For example, logarithms and square roots impose domain restrictions. A candidate is verified on an interval only when it is differentiable there and its derivative equals the integrand there.
Worked Example: Checking a Power of a Composite Function
Worked Example: Checking a Power of a Composite Function
Check whether \(F(x)=\frac{1}{3}(x^2+1)^{3/2}\) is an antiderivative of \(f(x)=x\sqrt{x^2+1}\) for all real \(x\).
The inner function \(x^2+1\) is positive for all real \(x\), so the candidate is differentiable on \(\mathbb{R}\). Apply the constant multiple rule and then the chain rule:
This is exactly \(f(x)\). Therefore \(F\) is an antiderivative of \(f\) on \(\mathbb{R}\). The final equality checks the candidate against the original integrand, not just against an intermediate expression.
Why Substitution-Based Candidates Need a Chain-Rule Check
Suppose a candidate has the form \(F(x)=H(g(x))\), where \(g\) is an inner function. The chain rule gives
Thus, if \(H'(u)=h(u)\), differentiating the composed candidate gives \(h(g(x))g'(x)\), not merely \(h(g(x))\). The extra factor \(g'(x)\) is precisely what must be present in the integrand for this candidate to work.
This also explains why substitution-based integration results can fail a check by a constant factor: a factor associated with the inner derivative may have been dropped or handled incorrectly.
Worked Example: Finding a Missing Factor
Worked Example: Finding a Missing Factor
A proposed antiderivative of \(f(x)=\frac{e^{3x}}{1+e^{3x}}\) is \(F(x)=\ln(1+e^{3x})\). Check it and correct it if necessary.
The logarithm's argument is positive for every real \(x\), so we may differentiate on \(\mathbb{R}\). For the proposed candidate, use the chain rule:
The proposed derivative is three times the target integrand, so the candidate is not correct. Multiplying the candidate by \(\frac{1}{3}\) corrects the discrepancy. Let \(G(x)=\frac{1}{3}\ln(1+e^{3x})\). Then
The corrected candidate passes the check. The missing factor was the reciprocal of the \(3\) produced by differentiating the inner function \(3x\).
Signs, Constants, and What the Derivative Cannot Detect
A sign error is visible in the same way as a factor error: differentiate the candidate and compare the result. An additive constant is different. The derivative of a constant is zero, so adding or removing a constant does not change whether a candidate is an antiderivative.
Here is why the last statement holds. Suppose \(F\) and \(G\) are differentiable on an interval \(I\) and both have derivative \(f\) there. Their difference \(D=F-G\) satisfies \(D'(x)=F'(x)-G'(x)=0\). For any \(a,b\in I\) with \(a<b\), the mean value theorem applies to \(D\) on \([a,b]\): \(D\) is continuous on that closed interval and differentiable in its interior. Therefore some \(c\in(a,b)\) satisfies
So \(D(b)=D(a)\) for any two points in \(I\); hence \(D\) is constant on \(I\). This is why checking by differentiation cannot determine the additive constant.
Worked Example: Checking a Logarithmic Candidate
Worked Example: Checking a Logarithmic Candidate
Check whether \(F(x)=\ln(x^2+4)\) is an antiderivative of \(f(x)=\frac{2x}{x^2+4}\), and compare it with \(G(x)=\frac{1}{2}\ln(x^2+4)\).
Since \(x^2+4>0\) for every real \(x\), both candidates are differentiable on \(\mathbb{R}\). Differentiate \(F\):
Thus \(F\) is verified. For the second candidate,
The second candidate has an incorrect multiplicative factor and is not an antiderivative of \(f\). By contrast, \(F(x)+7\) is valid because
The constant changes the candidate's values but not its derivative.
Worked Example: Checking a Candidate with a Square-Root Inner Function
Worked Example: Checking a Candidate with a Square-Root Inner Function
For \(x>0\), check whether \(F(x)=-2\cos(\sqrt{x})\) is an antiderivative of \(f(x)=\frac{\sin(\sqrt{x})}{\sqrt{x}}\).
The condition \(x>0\) ensures that \(\sqrt{x}\) is differentiable and the target integrand is defined. Applying the chain rule gives
The two factors of \(2\) cancel, and the candidate is verified on \((0,\infty)\). Omitting the derivative \(\frac{1}{2\sqrt{x}}\) of the inner function would miss the denominator in the target integrand.
Common Pitfall: Checking Only the Outer Function
A frequent error is to differentiate a composite expression as though its inner function were simply \(x\). For instance, the incorrect calculation
omits the derivative of \(x^2\). The chain rule instead gives
Consequently, if the target integrand is \(x\cos(x^2)\), the candidate \(\sin(x^2)\) is not correct: its derivative has twice the required value. The corrected candidate is \(\frac{1}{2}\sin(x^2)\), since
The practical correction is simple: when an expression contains an inner function, write its derivative explicitly before simplifying. This makes omitted chain-rule factors and incorrect signs easier to spot.
What a Successful Check Establishes
If a candidate is differentiable on an interval and its derivative agrees with the integrand at every point of that interval, it is an antiderivative there. A few matching sample values are not enough: the equality of derivatives must hold throughout the stated interval. Conversely, if the derivatives do not agree, the candidate is not an antiderivative of that integrand on that interval.
A check can also help locate the type of error. A constant ratio between the computed derivative and the target often signals a missing factor; an opposite sign signals a sign error; a missing or extra variable factor may point to a chain-rule mistake. These are diagnostic clues, but the decisive test remains direct differentiation and comparison.
Check Your Understanding
For each question, differentiate the proposed expression before deciding whether it passes the check.
- Is \(F(x)=\frac{1}{4}(x^2+1)^2\) an antiderivative of \(x(x^2+1)\) on \(\mathbb{R}\)?
- For \(x>0\), does \(F(x)=\ln(x^2)\) differentiate to \(\frac{2}{x}\)?
- If \(F'(x)=f(x)\), does \(F(x)-12\) also have derivative \(f(x)\)? Explain why.
- What chain-rule factor is missing if someone writes \(\frac{d}{dx}\cos(5x)= -\sin(5x)\)?
- For \(x>0\), check whether \(F(x)=\frac{2}{3}x^{3/2}\) is an antiderivative of \(f(x)=\sqrt{x}\).
