Why Test a Difference Quotient?
A graph may look smooth near a point, or a formula may appear familiar, but the definition of differentiability asks a precise question: does the function’s average rate of change over a shrinking interval approach one finite number? The difference quotient records that average rate of change. Testing its limit is the direct way to decide whether the derivative exists at the point.
The calculation has two essential parts. First, form the quotient using the function value at the point and a nearby function value. Then take the limit as the nearby input approaches the point, without setting the increment equal to zero. The quotient must approach a finite real number for the function to be differentiable there.
The condition that \(f\) is defined on an open interval containing \(a\) means inputs slightly smaller and slightly larger than \(a\) are in the domain. The quotient itself is used only for \(h\ne 0\), since division by zero is undefined. The limit asks what happens as such nonzero increments get arbitrarily close to zero.
What the Quotient Measures
For a nonzero increment \(h\), the inputs \(a\) and \(a+h\) are separated by \(h\). The function values change by \(f(a+h)-f(a)\). Their ratio is the average change in output per unit change in input across those two inputs. As \(h\) approaches zero, the quotient tests whether these average rates approach a single finite value.
The same definition is often written with \(x\) approaching \(a\):
provided the limit exists as a finite real number. To see why the two forms agree, put \(x=a+h\). Then \(x-a=h\), and \(x\to a\) exactly when \(h\to 0\). Thus the quotient in the \(x\)-form becomes the quotient in the increment form. This is a change of variable, not a different test.
A Reliable Test Procedure
For a function defined near \(a\), use the following sequence. The algebraic simplification is usually the main work: the original expression may look undefined when \(h=0\), even though its limit is straightforward.
Use \(\frac{f(a+h)-f(a)}{h}\), keeping \(h\ne0\).
Calculate \(f(a+h)\) and \(f(a)\) separately so the subtraction is clear.
Factor, combine terms, or rationalize when appropriate. Do not cancel a factor unless it is nonzero in the quotient’s domain.
Determine whether the quotient approaches one finite real number as \(h\to0\).
If that finite limit is \(L\), then \(f\) is differentiable at \(a\) and \(f'(a)=L\). If it does not exist as a finite number, this test does not give a derivative at \(a\).
Worked Examples
Worked Example: A Quadratic Function at a Point
Test whether \(f(x)=x^2+3x\) is differentiable at \(a=2\), and find the derivative if it exists.
First calculate the function value at the point:
Now calculate the nearby value:
Substitute into the difference quotient and simplify for \(h\ne0\):
Therefore,
This is finite, so \(f\) is differentiable at \(2\), with \(f'(2)=7\). Check the limiting result by subtracting the proposed value from the simplified quotient: \((7+h)-7=h\), and \(\lim_{h\to0}h=0\). The quotient does approach \(7\).
Worked Example: Rationalizing a Square-Root Quotient
Let \(f(x)=\sqrt{x}\). Test differentiability at \(a=4\).
Since \(f(4)=2\) and \(f(4+h)=\sqrt{4+h}\), the quotient is
For nonzero \(h\) near zero, multiply numerator and denominator by the conjugate \(\sqrt{4+h}+2\):
The cancellation is valid because \(h\ne0\) in the difference quotient. Also, \(4+h\) remains positive for all sufficiently small \(h\). Thus
The limit is finite, so \(f\) is differentiable at \(4\) and \(f'(4)=\frac14\). For a direct check of the limit, the simplified quotient minus \(\frac14\) is \[ \frac{1}{\sqrt{4+h}+2}-\frac14 =\frac{2-\sqrt{4+h}}{4(\sqrt{4+h}+2)}. \] The numerator approaches \(0\), while the denominator approaches \(16\), so this difference approaches \(0\).
Worked Example: A Reciprocal Function at a Nonzero Input
Let \(f(x)=\frac1x\). Test differentiability at \(a=2\).
We have \(f(2)=\frac12\) and \(f(2+h)=\frac{1}{2+h}\). For \(h\) sufficiently close to zero, \(2+h\ne0\), and the difference quotient is
Here \(h\ne0\) allows cancellation, and \(2+h\ne0\) near zero. Taking the limit gives
The quotient has a finite limit, so \(f\) is differentiable at \(2\), with \(f'(2)=-\frac14\). To check the value, compare the simplified quotient with \(-\frac14\): \[ -\frac{1}{2(2+h)}+\frac14=\frac{h}{4(2+h)}. \] As \(h\to0\), this difference tends to \(0\), confirming that the quotient approaches \(-\frac14\).
Why a Finite Limit Is Required
The definition does not say merely that the quotient has some behavior near zero. It requires the quotient to approach a particular real number. If the quotient becomes unbounded or fails to settle toward one value, there is no finite derivative supplied by the definition. This condition is part of the meaning of differentiability, not an optional preference about how to report an answer.
Worked Example: A Quotient That Does Not Have a Finite Limit
Define \(f(0)=0\), and for \(x\ne0\), define \(f(x)=x\sin(1/x^2)\). Test differentiability at \(0\).
For \(h\ne0\),
This expression does not approach a single value as \(h\to0\). To verify that, choose two sequences of nonzero increments:
As \(n\) increases without bound, both \(h_n\) and \(k_n\) approach \(0\). But the quotient along the first sequence is
while along the second it is
If the limit existed, values along every sequence approaching zero would approach the same limit. These two sequences instead give the constant values \(1\) and \(-1\). Therefore the difference quotient has no limit, and \(f\) is not differentiable at \(0\).
A Common Mistake: Forgetting to Subtract \(f(a)\)
The numerator is the change in the function value, not just the nearby function value. Omitting \(f(a)\) changes the expression being tested and can give a completely different result.
What to Conclude from the Test
When simplifying, keep track of the original restriction \(h\ne0\). It is legitimate to cancel a factor of \(h\) in the quotient because the limit considers nonzero increments; it is not legitimate to claim that the original quotient is defined at \(h=0\). After simplification, evaluate the limit as \(h\) approaches zero.
If the resulting limit is a finite number \(L\), the definition gives both conclusions: the function is differentiable at \(a\), and \(f'(a)=L\). If the limit fails to exist or does not approach a finite real value, the function is not differentiable at \(a\) according to this definition. The test is local: it uses the function’s values at \(a\) and at inputs arbitrarily close to \(a\).
On the AP exam, showing the quotient and its limiting behavior makes the reasoning visible. A final number alone does not establish differentiability; the work should make clear that the number comes from the limit of \(\frac{f(a+h)-f(a)}{h}\).
Check Your Understanding
For each question, begin with the difference quotient and explain what its limit tells you.
- For \(f(x)=x^2-4x\), test differentiability at \(a=1\) and find \(f'(1)\), if it exists.
- For \(g(x)=\sqrt{x}\), write the difference quotient at \(a=9\), rationalize it, and determine whether the limit is finite.
- Why must the quotient be formed using \(f(a+h)-f(a)\), rather than \(f(a+h)\) alone?
- What does a finite limit \(L\) of the difference quotient establish about \(f\) at \(a\)?
- If two sequences of nonzero increments both approach zero but the corresponding quotient values approach different numbers, can the derivative exist? Explain using the limit definition.
