Why a Constant Has Zero Derivative
A derivative measures how a function changes as its input changes. If the function always returns the same value, changing the input produces no change in its output. The difference quotient makes this idea precise: for a constant function, the change in output is zero for every allowed nonzero change in input.
This result is a basic differentiation rule, but it follows directly from the definition of the derivative. It also explains what happens to a constant term in a formula: the term contributes no change to the numerator of the difference quotient.
Definition and Difference Quotient
For a function defined on an open interval containing \(a\), its derivative at \(a\), when the limit exists, is defined by the difference quotient:
Here \(h\) is a nonzero change in the input, small enough that \(a+h\) remains in the domain. The numerator \(f(a+h)-f(a)\) is the corresponding change in the output. The quotient compares those two changes; the limit describes the resulting rate as the input change approaches zero.
The Derivative of a Constant Function
Suppose \(f(x)=c\) throughout an open interval, and let \(a\) be any point in that interval. For every sufficiently small nonzero \(h\), both \(a\) and \(a+h\) are in the domain. Since the function has the same value at both inputs, the difference quotient simplifies completely.
This quotient is exactly zero for every allowed nonzero \(h\), not merely close to zero. Therefore, its limit as \(h\) approaches zero is zero.
The condition that the derivative is considered at an interior point ensures that small changes in either direction stay within the domain. The conclusion concerns the derivative at points where the derivative is defined; it does not assert a two-sided derivative at an endpoint of a domain.
Worked Examples
Worked Example: A Constant Function with Value 5
Let \(f(x)=5\) for all real \(x\). Find \(f'(a)\) using the difference quotient, where \(a\) is any real number.
First evaluate the function at the two inputs in the quotient: \(f(a+h)=5\) and \(f(a)=5\). For \(h\ne 0\),
The quotient is zero for every nonzero \(h\), so its limit as \(h\to 0\) is zero. Thus \(f'(a)=0\) for every real \(a\).
Check: Substituting the proposed derivative into the definition gives a difference quotient equal to \(0\) for every allowed \(h\), as required.
Worked Example: A Constant Position
Suppose an object remains at position \(s(t)=12\) metres throughout a time interval, where \(t\) is measured in seconds. Find its derivative at an interior time \(t=a\).
For a nonzero time change \(h\) that keeps both times in the interval, the positions are \(s(a+h)=12\) metres and \(s(a)=12\) metres. The difference quotient is
Since this value is zero for every allowed nonzero \(h\), the derivative is \(s'(a)=0\text{ m/s}\). The object’s position is 12 metres, but its rate of change of position is zero metres per second.
Check: In the difference quotient, the output change is \(12\text{ m}-12\text{ m}=0\text{ m}\), so the quotient is \(0\text{ m/s}\) for every allowed \(h\). This agrees with the derivative found.
Worked Example: A Constant Term Added to a Function
Suppose \(g\) is differentiable at \(a\), with \(g'(a)=2\), and define \(f(x)=g(x)-6\) near \(a\). Find \(f'(a)\) by showing exactly what happens to the constant term.
For every sufficiently small nonzero \(h\), expand the difference quotient using the definition of \(f\):
The \(-6\) and \(+6\) cancel, so the quotient for \(f\) is exactly the quotient for \(g\). Taking limits as \(h\to 0\) gives \(f'(a)=g'(a)=2\).
Check: The definition of \(g'(a)=2\) says that the limit of \(\frac{g(a+h)-g(a)}{h}\) is 2. The displayed equality shows the quotient for \(f\) has the same limit, confirming \(f'(a)=2\).
Why the Constant Cancels
The cancellation is not a special trick for the number 6. If \(C\) is any fixed constant and \(f(x)=g(x)+C\), then
whenever the expressions are defined and \(h\ne 0\). Thus, if \(g\) is differentiable at \(a\), \(f\) is differentiable there and has the same derivative. The reasoning depends on an exact equality of difference quotients, so their limits are equal.
This explains the role of an additive constant in a formula. A fixed term has the same value at the two inputs, so its contribution to the output change is the difference \(C-C=0\). The result is about the constant term’s contribution: it does not by itself calculate the derivative of the remaining, possibly nonconstant, part.
Common Pitfall: Confusing Function Value and Derivative
A constant function can have a nonzero value even though its derivative is zero. For example, if \(f(x)=5\), then the statement \(f'(x)=5\) is incorrect. It confuses the function value, which is 5, with the derivative, which measures change.
One way to see the error is to test the incorrect answer against the definition. The difference quotient for \(f(x)=5\) is
Its limit is 0, not 5. The corrected conclusion is \(f'(x)=0\) at every interior point of the domain. A nonzero constant function is still constant; the height of its graph does not imply that the graph is changing.
Why the Result Matters
Many formulas include fixed terms, such as an initial amount, a baseline measurement, or a constant offset. When differentiating such a formula, the difference quotient compares the formula’s values at two inputs. A fixed additive term appears unchanged in both values and cancels when they are subtracted.
This observation is useful before applying any broader differentiation rules. It lets you identify the constant part without mistaking its size for a rate of change. The derivative of a constant function is always zero; when a constant is added to a differentiable function, that addition leaves the derivative unchanged.
Check Your Understanding
Use the difference quotient and the meaning of a constant term to answer these questions.
- For \(f(x)=-3\), simplify \(\frac{f(a+h)-f(a)}{h}\) for \(h\ne 0\), then state \(f'(a)\).
- Why must \(a+h\) remain in the domain when forming the difference quotient at \(a\)?
- If \(f(x)=g(x)+9\) and \(g\) is differentiable at \(a\), explain why the difference quotients for \(f\) and \(g\) are equal.
- A function has the constant value 20. Is its derivative 20 or 0? Explain the difference between the function value and its derivative.
- What domain condition ensures the usual two-sided derivative calculation at \(a\) can use sufficiently small changes in both directions?
