Why One-Sided Derivatives Matter
A derivative at a point describes the limit of nearby difference quotients. When the point lies inside the function’s domain, nearby inputs can approach from either side. The derivative exists only if those approaches produce one common, finite value.
Separating the left-hand and right-hand limits makes this test precise. It also prevents an important mistake: calculating a limit from one side and treating it as if it automatically describes the behavior from both sides.
Definitions
Let \(f\) be defined at \(a\), and suppose its domain contains points arbitrarily close to \(a\) on the left. For \(x\ne a\), the difference quotient at \(a\) is \(\frac{f(x)-f(a)}{x-a}\). If its limit as \(x\) approaches \(a\) from the left exists as a finite real number, that limit is the left-hand derivative.
Equivalently, using the increment \(h=x-a\), approaching \(a\) from the left means \(h\to 0^-\):
For the right-hand derivative, the domain must contain points arbitrarily close to \(a\) on the right, and the increment approaches zero through positive values:
These definitions require the function value \(f(a)\). They also require the relevant nearby domain points: a left-hand derivative requires points in the domain to the left, while a right-hand derivative requires points in the domain to the right. For an ordinary two-sided derivative, points from both sides are needed.
Worked Example: Both One-Sided Derivatives of a Quadratic
Worked Example: A Two-Sided Derivative at \(a=2\)
Let \(f(x)=x^2\). Calculate the left- and right-hand derivatives at \(a=2\), then decide whether \(f'(2)\) exists.
First, \(f(2)=2^2=4\). The difference quotient is
For the left-hand limit, let \(h\to0^-\). Since \(4+h\to4\),
For the right-hand limit, let \(h\to0^+\). The same simplified expression tends to the same value:
Both one-sided derivatives are finite and equal. Therefore the two-sided derivative exists and \(f'(2)=4\). As a check, both calculations used the original quotient with \(f(2)=4\); after cancellation its value is \(4+h\), which approaches \(4\) from either side.
The Two-Sided Criterion
The reason is the corresponding criterion for ordinary two-sided limits. A difference quotient has a finite two-sided limit exactly when its left-hand and right-hand limits both exist as finite numbers and agree. Here the expression whose limits are compared is \(\frac{f(x)-f(a)}{x-a}\).
For \(x\ne a\), define \(Q(x)=\frac{f(x)-f(a)}{x-a}\). The derivative is the limit of \(Q(x)\) as \(x\to a\).
Values of \(x\) near \(a\) are either to its left or to its right. Thus a two-sided limit must agree with the limits of \(Q(x)\) on both sides.
If the left-hand and right-hand limits both equal the same finite number \(L\), then \(Q(x)\to L\) from either side, so \(f'(a)=L\). If either limit fails to exist as a finite number, or if they differ, there is no finite two-sided limit.
Why the Criterion Works
Here is a direct justification using the definition of a limit. Suppose the left-hand and right-hand derivatives both exist and equal \(L\). Given any \(\varepsilon>0\), the left-hand limit gives a distance \(\delta_1>0\) such that
The right-hand limit gives a distance \(\delta_2>0\) such that
Choose \(\delta=\min(\delta_1,\delta_2)\). If \(0<|x-a|<\delta\), then either \(x<a\) or \(x>a\). In the first case the left-hand condition applies; in the second case the right-hand condition applies. In both cases the quotient differs from \(L\) by less than \(\varepsilon\). Therefore its two-sided limit is \(L\), so \(f'(a)=L\).
Conversely, suppose the two-sided derivative exists and equals \(L\). By the definition of a two-sided limit, the quotient approaches \(L\) whenever \(x\) approaches \(a\) with \(x\ne a\). Restricting that same condition to \(x<a\) gives the left-hand limit \(L\); restricting it to \(x>a\) gives the right-hand limit \(L\). Thus both one-sided derivatives exist, are finite, and equal the two-sided derivative. This proves both directions of the criterion.
Worked Example: Matching One-Sided Derivatives
Worked Example: Two Formulas with the Same Derivative at the Join
Define \(f\) by \(f(x)=x^2\) for \(x\le2\), and \(f(x)=x^2+(x-2)^2\) for \(x>2\). Find both one-sided derivatives at \(a=2\).
At the point, the first rule gives \(f(2)=2^2=4\). For \(h<0\), \(2+h<2\), so the left-hand quotient uses \(f(2+h)=(2+h)^2\):
Therefore \(f'_-(2)=\lim_{h\to0^-}(4+h)=4\).
For \(h>0\), \(2+h>2\), so the second rule applies. Substituting into it gives
Consequently, the right-hand quotient is
Thus \(f'_+(2)=\lim_{h\to0^+}(4+2h)=4\). The two finite one-sided values agree, so \(f'(2)=4\). Checking the result directly, the left-hand quotient tends to \(4\), and the right-hand quotient also tends to \(4\), as required.
Worked Example: Unequal One-Sided Derivatives
Worked Example: The Two-Sided Derivative Does Not Exist
Let \(g(x)=x^2\) for \(x\le0\), and \(g(x)=2x^2+x\) for \(x>0\). Determine whether \(g'(0)\) exists.
First, \(g(0)=0^2=0\). If \(h<0\), then \(g(h)=h^2\), so
Taking the left-hand limit gives \(g'_-(0)=\lim_{h\to0^-}h=0\).
If \(h>0\), then \(g(h)=2h^2+h\). The quotient is
Taking the right-hand limit gives \(g'_+(0)=\lim_{h\to0^+}(2h+1)=1\). Both one-sided derivatives are finite, but \(0\ne1\). By the criterion, \(g'(0)\) does not exist. Checking the conclusion against the two quotients confirms the mismatch: the left quotient tends to \(0\), while the right quotient tends to \(1\).
Common Mistake: Ignoring the Direction of Approach
A common error is to calculate one one-sided limit and label it as the derivative without checking the other side. For the function \(g\) in the preceding example, the calculation \(\lim_{h\to0^-}h=0\) correctly gives \(g'_-(0)=0\). It does not establish \(g'(0)=0\), because it says nothing about positive \(h\). The right-hand quotient is \(2h+1\), whose limit is \(1\). Since the two limits differ, the two-sided derivative does not exist.
Another error is to substitute \(h=0\) into the difference quotient before simplifying. The quotient has denominator \(h\), so direct substitution produces division by zero. Instead, simplify the quotient for \(h\ne0\), then take the appropriate one-sided limit.
Check Your Understanding
For each question, use the difference quotient and state whether the two-sided derivative exists when requested.
- For \(f(x)=x^2\), write the difference quotient at \(a=3\) using \(h\), then find \(f'_-(3)\) and \(f'_+(3)\).
- A function has \(f'_-(a)=5\) and \(f'_+(a)=5\). What is \(f'(a)\), assuming it is defined on both sides of \(a\)?
- A function has \(f'_-(a)=-2\) and \(f'_+(a)=4\). Does its two-sided derivative exist?
- In the definition of \(f'_+(a)\), should \(h\) approach zero through positive or negative values?
- Why does calculating only \(f'_-(a)\) not establish that \(f'(a)\) exists?
