One Derivative, Many Functions
A derivative describes how a function changes, but it does not record the function's vertical position. If a graph is shifted up or down by a fixed amount, its rate of change at every input stays the same.
This means that when we reverse the question “What is the derivative of this function?”, we should not expect a single original function. Instead, we look for a family of functions with the same derivative. The key is to define antiderivatives precisely and use differentiation to check which functions belong to that family.
Definition of an Antiderivative
Let \(f\) be a function defined on an open interval \(I\). A function \(F\) is an antiderivative of \(f\) on \(I\) if \(F\) is differentiable there and its derivative equals \(f\) at every input in the interval.
The condition must hold throughout the stated interval, not just at one input or at a few tested values. The direct way to verify a proposed antiderivative is to differentiate it and compare the result with \(f\).
Worked Example: Checking a Proposed Antiderivative
Determine whether \(F(x)=x^2+4\) is an antiderivative of \(f(x)=2x\) on \(\mathbb{R}\).
Differentiate each term of \(F\). The derivative of \(x^2\) is \(2x\), and the derivative of the constant \(4\) is \(0\). Thus
Since \(2x=f(x)\) for every real \(x\), the definition is satisfied. Therefore \(F\) is an antiderivative of \(f\) on \(\mathbb{R}\). As a check, substituting any input gives matching values for \(F'(x)\) and \(f(x)\); for example, at \(x=3\), both equal \(6\).
Vertical Shifts Preserve Derivatives
Suppose \(F\) is an antiderivative of \(f\), and let \(C\) be any real number that does not depend on \(x\). Differentiating \(F(x)+C\) gives
So every function obtained by adding a constant to \(F\) is also an antiderivative of \(f\). Geometrically, adding \(C\) shifts the graph vertically. The slope at each input is unchanged, so the derivative is unchanged as well.
Worked Example: A Family with the Same Derivative
Start with \(F(x)=3x^2-4x\), and consider the functions \(G(x)=3x^2-4x+7\) and \(H(x)=3x^2-4x-2\). Find their derivatives and compare them.
For \(F\), differentiate each term:
For \(G\), the added term \(7\) is constant, so
For \(H\), the added term \(-2\) is also constant, so
All three functions have derivative \(6x-4\), so they are antiderivatives of the same function. Their differences confirm the vertical shifts: \(G(x)-F(x)=7\) and \(H(x)-F(x)=-2\). For a numerical check at \(x=1\), each derivative equals \(6(1)-4=2\).
Why These Are All the Antiderivatives
Adding a constant always produces another antiderivative. There is also a converse: on an interval, any two antiderivatives of the same function differ by a constant. The interval condition matters because it allows us to apply the Mean Value Theorem between any two inputs in the domain.
To prove the claim, subtract the functions and define \(H(x)=F(x)-G(x)\). Differentiating gives \(H'(x)=F'(x)-G'(x)=0\) throughout \(I\). Take any two points \(a,b\in I\) with \(a<b\). Since \(I\) is an interval, \([a,b]\) lies within \(I\). The function \(H\) is continuous on \([a,b]\) and differentiable on \((a,b)\), so the Mean Value Theorem applies. It guarantees some \(c\in(a,b)\) such that
Because \(H'(c)=0\), this equation gives \(H(b)-H(a)=0\), or \(H(b)=H(a)\). The two points were arbitrary, so \(H\) has the same value at every input in \(I\). Call that value \(C\). Then \(F(x)-G(x)=C\), and therefore \(F(x)=G(x)+C\) throughout \(I\). This proves that no other differences between antiderivatives are possible.
If \(F'=G'\), then the derivative of their difference is zero.
A function with zero derivative throughout an interval has the same value at any two points of that interval.
If \(F-G\) has the same value everywhere, \(F\) and \(G\) differ by a constant.
Using Derivatives to Compare Candidates
The theorem gives a reliable test: differentiate each candidate. If both derivatives equal the given function, they belong to the same family on the interval. If one derivative is different, that candidate does not meet the definition.
Worked Example: Rejecting a Candidate That Changes with x
For \(f(x)=2x+2\), check whether \(F(x)=x^2+2x+3\) and \(G(x)=x^2+5x+3\) are both antiderivatives.
First differentiate \(F\):
This equals \(f(x)\), so \(F\) is an antiderivative. Next differentiate \(G\):
This is not equal to \(2x+2\). Their derivative difference is \(3\), not \(0\), so \(G\) is not an antiderivative of \(f\). The function difference also shows why: \(G(x)-F(x)=3x\), which is not constant. For a check at \(x=0\), \(f(0)=2\), \(F'(0)=2\), and \(G'(0)=5\); the first candidate matches and the second does not.
A Common Mistake
A frequent error is to differentiate a function and keep its added constant as if it still affected the derivative. For example, the incorrect calculation
treats \(7\) as though it changes with \(x\). But \(7\) has the same value for every input, so its derivative is \(0\). The corrected calculation is
Keep separate the function value and its rate of change: a constant can change the function's vertical position without changing any of its slopes.
Why the Family View Matters
When a problem asks for an antiderivative, a single verified function is useful, but it does not describe every possible answer. The family viewpoint records that all vertical shifts have the same derivative. If another condition later specifies a value of the function, that extra information can select one member of the family; without such a condition, the derivative alone does not determine the vertical position.
The interval qualification also prevents an overgeneralization. The theorem says that two antiderivatives on one interval differ by one constant throughout that interval. If a domain is split into separate intervals, the comparison on one interval does not by itself require the same constant on another. Always identify the domain when describing the family.
Check Your Understanding
Use differentiation and the interval condition in your answers.
- Is \(x^2-6\) an antiderivative of \(2x\) on \(\mathbb{R}\)? Show the derivative check.
- If \(F\) is an antiderivative of \(f\), why is \(F(x)+12\) also an antiderivative of \(f\)?
- Suppose \(F'\) and \(G'\) are equal throughout an open interval. What can you conclude about \(F-G\), and which theorem justifies the conclusion?
- For \(f(x)=2x+2\), is \(x^2+2x-4\) an antiderivative? Verify your answer.
- Why does the result about two antiderivatives being separated by a constant require attention to the domain?
