One Point Is Not an Entire Interval
A derivative is first defined at a particular input. To say that a function is differentiable at \(a\) is to make a claim about the function’s behavior near that one input. To say that it is differentiable on an interval is a stronger claim: the derivative must exist at every point in that interval, with endpoint conventions stated when needed.
This distinction matters in practice. A difference quotient can have a limit at one chosen point even if the function has a problem elsewhere. A formula may also be differentiable wherever it is defined but fail to be a function on an interval that includes an excluded input. The interval and the domain must therefore be checked along with the limit.
The condition \(a+h\in A\) is essential: the quotient can only use inputs where \(f\) is defined. When \(a\) is an endpoint of the domain, the limit is taken from the side that stays in the domain, as in the usual one-sided endpoint convention.
For an open interval \(I\), \(f\) is differentiable on \(I\) if it is defined at every point of \(I\) and differentiable at every \(a\in I\). For a closed interval, we will include one-sided derivatives at the endpoints when saying it is differentiable throughout the interval. In every case, the requirement is universal: no point in the stated interval may be omitted.
A Pointwise Derivative of a Polynomial
Begin with a polynomial, for which the domain is all real numbers. We can calculate a derivative at one point directly from the definition, then compare that calculation with the stronger calculation at an arbitrary input.
Worked Example: A Cubic at One Point and Throughout the Real Line
Let \(p(x)=x^3-2x+4\). First find the derivative at \(a=1\) from the difference quotient. The function value is
At the nearby input \(1+h\),
Thus, for \(h\ne 0\),
Taking the limit gives
To show differentiability at every real input, let \(a\in\mathbb{R}\) be arbitrary. Expand the numerator:
For \(h\ne 0\), divide by \(h\) and take the limit:
This limit exists for every real \(a\), so \(p\) is differentiable on \(\mathbb{R}\). Substituting \(a=1\) into the general result gives \(p'(1)=3(1)^2-2=1\), agreeing with the pointwise calculation.
The first calculation established just one value, \(p'(1)\). The second established a result for an arbitrary real input, and therefore supports the claim that the polynomial is differentiable throughout \(\mathbb{R}\). Moving from a particular point to an arbitrary one is not just a longer calculation: it changes the scope of the conclusion.
From a Pointwise Limit to an Interval Claim
The logic can be stated precisely. Differentiability at a specified \(a\) asks whether one limit exists. Differentiability on an open interval \(I\) asks whether that limit exists for every input in \(I\), and whether the function is defined throughout \(I\).
In the limit at each \(a\), \(h\) must be small enough that \(a+h\) remains in the domain. This requirement is automatic for an interior point of an open interval on which \(f\) is defined. It is not automatic for a formula whose domain has a hole or a forbidden input.
One useful way to justify a claim on an interval is to take an arbitrary point of that interval and show that its difference quotient has a finite limit. If the calculation works for every such point, it proves the interval claim. A calculation at one selected point, however accurate, cannot substitute for that argument.
Rational Functions and Their Domains
A rational formula involves a quotient of polynomials, so its denominator restricts where the function is defined. A difference-quotient calculation can establish differentiability at an input only if the function is defined there and at the nearby inputs used in the limit. Even if algebra simplifies a formula, that simplification does not by itself change the original domain.
Worked Example: A Rational Function on Each Side of an Excluded Input
Consider
The input \(2\) is excluded because it makes the denominator zero. First calculate the derivative at \(a=0\). Since \(r(0)=1/(-2)=-\frac12\), for nonzero \(h\) near zero we have
Therefore,
Now take any \(a\ne 2\). Polynomial division gives \(x^2+1=(x-2)(x+2)+5\), so on the domain of \(r\),
For \(h\ne 0\) with \(a+h\ne 2\), the difference quotient is
Because \(a\ne 2\), the denominator \(a-2\) is nonzero. For \(h\) sufficiently close to zero, \(a+h-2\) also stays nonzero, and the limit is
This proves differentiability at every \(a\ne 2\). In particular, at \(a=0\), the formula gives \(1-\frac{5}{4}=-\frac14\), which checks the earlier calculation. The function is differentiable on each interval \((-\infty,2)\) and \((2,\infty)\), but it is not a function on an interval containing \(2\).
The final qualification is important. The derivative formula is valid only for \(a\ne 2\), just as the original rational function is defined only there. It would be incorrect to use the formula to claim differentiability at the excluded input.
A Pointwise Derivative Despite a Hole Elsewhere
The next example makes the distinction especially visible: a rational function has a derivative at a chosen point, yet it cannot be differentiable throughout an interval containing a different, excluded input.
Worked Example: Differentiable at Zero but Not Defined on the Whole Interval
Let
At \(x=0\), the function is defined and \(q(0)=(-1)/(-1)=1\). For \(h\) sufficiently close to zero, \(h\ne 1\), so cancellation is valid at that input:
Consequently, for nonzero \(h\) sufficiently close to zero,
The difference quotient is constantly \(1\) for these \(h\), so \(q'(0)=1\). Nevertheless, \(q\) is not differentiable throughout \((-1,2)\): it is not even defined at \(x=1\), which belongs to that interval. Thus differentiability at zero is true, while the claim of differentiability on the whole interval is false.
Common Pitfall: Cancelling Does Not Add an Input to the Domain
A tempting but incorrect argument in the preceding example is: “The expression simplifies to \(x+1\), so the function is defined and differentiable at \(x=1\).” The algebraic cancellation is valid only when \(x\ne 1\). In fact, substituting \(x=1\) into the original formula gives a zero denominator, so the original function has no value there.
The corrected conclusion is that \(q(x)=x+1\) on its domain \(\mathbb{R}\setminus\{1\}\), and the derivative at \(0\) exists. The cancellation does not make the original function defined at \(1\), so it cannot establish differentiability on any interval that includes \(1\). Keep the formula and its domain together whenever assessing an interval claim.
A Reliable Check for Interval Differentiability
When deciding whether a function is differentiable throughout an interval, separate the task into two questions. First check that every point of the interval belongs to the function’s domain. Then check the derivative limit at an arbitrary point of the interval, including the appropriate one-sided limit if it is an endpoint. A failure at even one point disproves the full interval claim.
Identify excluded inputs, such as zeros of a rational function’s denominator. If the interval contains an input where the function is undefined, it cannot be differentiable throughout that interval.
Use the difference quotient at a specified point for a pointwise claim. For an interval claim, let the input be arbitrary and establish the limit for every point in the interval.
A limit at one input proves differentiability there. A result valid at every input, together with the required domain condition, supports differentiability throughout the interval.
This approach prevents two common overstatements: inferring differentiability everywhere from one successful limit, and treating an algebraically simplified expression as if it had changed the original domain. A pointwise result is valuable, but it answers only the pointwise question unless the reasoning covers all points under consideration.
Check Your Understanding
For each question, pay attention to whether the claim concerns one input or every input in an interval.
- State the difference-quotient definition of differentiability at an interior point \(a\).
- What additional requirement is needed to claim that a function is differentiable on an open interval?
- For \(r(x)=\frac{x^2+1}{x-2}\), why can one not claim differentiability on an interval containing \(2\)?
- In the example \(q(x)=\frac{x^2-1}{x-1}\), why does the simplification \(q(x)=x+1\) not define \(q(1)\)?
- If a difference quotient has a finite limit at \(a=0\), what does that establish, and what does it not establish about an interval containing zero?
