Why Examine the Two Sides Separately?
The derivative at a point measures the limiting change in a function as the input changes by a small amount. At a point where a function is defined by different formulas on the two sides, those changes may behave differently depending on the direction of approach. One-sided difference quotient limits let us examine each direction precisely.
The increment form of the difference quotient uses an increment \(h\):
For \(h>0\), the input \(a+h\) lies to the right of \(a\). For \(h<0\), it lies to the left. The numerator and denominator keep the same order in both cases; only the allowed direction of \(h\) changes.
Definitions of the One-Sided Derivatives
Assume that \(f(a)\) is defined and that \(f\) is defined on some interval immediately to the left and immediately to the right of \(a\). The left and right derivatives are the corresponding one-sided limits of the difference quotient, provided those limits exist as finite real numbers.
The notation \(h\to 0^-\) means that \(h\) approaches zero through negative values. The notation \(h\to 0^+\) means that it approaches through positive values. Equivalently, \(f'_{-}(a)\) uses inputs approaching \(a\) from the left, while \(f'_{+}(a)\) uses inputs approaching from the right.
The ordinary derivative uses the unrestricted two-sided limit:
For this two-sided derivative to be defined in the usual sense, \(a\) must be an interior point of the domain: the function must be defined on both sides of \(a\) in some interval around it. The ordinary derivative is a finite real number. Thus, a limit that grows without bound on one side is not a one-sided derivative under this definition.
Worked Example: Equal One-Sided Derivatives
Worked Example: Equal One-Sided Derivatives
Let \(f(x)=x^2\) for \(x<1\), and \(f(x)=2x-1\) for \(x\geq 1\). Find the left and right derivatives at \(a=1\), and determine whether \(f'(1)\) exists.
Since \(1\) belongs to the second branch, \(f(1)=2(1)-1=1\). For the left derivative, take \(h<0\). Then \(1+h<1\), so the left branch applies:
Therefore \(f'_{-}(1)=\lim_{h\to0^-}(2+h)=2\). For the right derivative, take \(h>0\). Then \(1+h\geq1\), so the second branch applies:
Thus \(f'_{+}(1)=2\). Both one-sided limits are finite and equal, so the two-sided derivative exists and \(f'(1)=2\). As a direct check, the left quotient approaches \(2\), and the right quotient is exactly \(2\) for every \(h>0\) close to zero.
Worked Example: Unequal One-Sided Derivatives
Worked Example: Unequal One-Sided Derivatives
Now let \(g(x)=x^2\) for \(x\leq1\), and \(g(x)=3x-2\) for \(x>1\). Determine the one-sided derivatives at \(1\) and whether the ordinary derivative exists.
The first branch includes \(x=1\), so \(g(1)=1^2=1\). If \(h<0\), then \(1+h\leq1\), and the first branch gives
It follows that \(g'_{-}(1)=2\). If \(h>0\), then \(1+h>1\), so the second branch applies:
Therefore \(g'_{+}(1)=3\). The one-sided limits are finite, but they are not equal. The two-sided difference quotient limit consequently does not exist, so \(g'(1)\) does not exist. The calculation checks the conclusion directly: approaching from the left gives slopes tending to \(2\), while approaching from the right gives slopes equal to \(3\).
Worked Example: A One-Sided Quotient That Is Not Finite
Worked Example: A One-Sided Quotient That Is Not Finite
Define \(q(x)=x+1\) for \(x<0\), and \(q(x)=x^2\) for \(x\geq0\). Determine what happens to the difference quotient on each side of \(a=0\).
The second branch gives \(q(0)=0^2=0\). When \(h<0\), the first branch applies, so
As \(h\to0^-\), \(1/h\to-\infty\). Hence the left difference quotient does not approach a finite real number, and \(q'_{-}(0)\) does not exist as a finite one-sided derivative. When \(h>0\), the second branch applies:
Thus the right derivative exists and equals \(q'_{+}(0)=0\), but there is no finite two-sided derivative because the left quotient has no finite limit. The difference quotient calculations, rather than any assumption about the shape of the graph, establish this result.
When Do the One-Sided Limits Give a Derivative?
This criterion follows from the basic relationship between two-sided and one-sided limits. Here is a proof for the difference quotient. Write \(Q(h)=\frac{f(a+h)-f(a)}{h}\) for \(h\ne0\) close to zero.
Suppose \(\lim_{h\to0}Q(h)=L\). Given any \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<|h|<\delta\) implies \(|Q(h)-L|<\varepsilon\). This implication holds in particular when \(-\delta<h<0\), and also when \(0<h<\delta\). Therefore both one-sided limits equal \(L\).
Suppose the left and right limits both exist and equal the finite number \(L\). For a given \(\varepsilon>0\), choose \(\delta_{-}>0\) so that \(-\delta_{-}<h<0\) implies \(|Q(h)-L|<\varepsilon\). Choose \(\delta_{+}>0\) so that \(0<h<\delta_{+}\) implies the same inequality. Set \(\delta=\min(\delta_{-},\delta_{+})\). Then every \(h\) with \(0<|h|<\delta\) is either negative or positive, and in either case \(|Q(h)-L|<\varepsilon\). This is the definition of \(\lim_{h\to0}Q(h)=L\).
The proof also shows why checking just one side is insufficient: a two-sided limit makes a demand about every sufficiently small nonzero \(h\), both negative and positive. If either side fails to approach a finite number, or if the two sides approach different numbers, that demand cannot be met.
A Common Sign Error
The increment quotient has a fixed order: \(f(a+h)-f(a)\) in the numerator and \(h\) in the denominator. The fact that \(h\) is negative on the left does not reverse the numerator. For the function \(g\) in the second worked example, the correct left quotient is
A common incorrect calculation swaps the function values but leaves the denominator unchanged:
This expression tends to \(-2\), but it is not the difference quotient defining the derivative: its numerator has been reversed without also reversing the denominator. The correction is to keep the prescribed order throughout. A negative increment already accounts for approaching from the left.
A Reliable Procedure
For a piecewise function at a point where the formula may change, proceed in a fixed order:
- Find \(f(a)\) from the branch that includes \(a\).
- Write the difference quotient \(\frac{f(a+h)-f(a)}{h}\).
- For the left limit, use the branch valid when \(h<0\); simplify before taking the limit.
- For the right limit, use the branch valid when \(h>0\); simplify before taking the limit.
- Check that each one-sided limit is finite, then compare their values.
The branch containing the point matters because it determines \(f(a)\), which appears in both quotients. The branch used for \(f(a+h)\) depends on the sign of \(h\). Keeping those two choices distinct prevents many errors in piecewise calculations.
Why the Comparison Matters
The one-sided calculations give a precise answer to a local question: does the difference quotient settle on one finite value regardless of the direction from which the input approaches \(a\)? If both sides settle on the same value, that value is the derivative. If they settle on different values, or if either side has no finite limit, the ordinary derivative does not exist.
This method is especially useful at points where a piecewise definition changes. It does not require guessing a tangent slope from a graph. Instead, it applies the derivative definition separately to negative and positive increments, and then uses the two-sided limit criterion to reach a conclusion.
Check Your Understanding
Use the difference quotient and keep track of which branch applies on each side.
- In the definition of \(f'_{-}(a)\), what restriction is placed on \(h\)?
- For \(r(x)=x^2\) when \(x<2\) and \(r(x)=4x-4\) when \(x\geq2\), calculate \(r'_{-}(2)\) and \(r'_{+}(2)\). Does \(r'(2)\) exist?
- Why must \(f(a)\) be found before calculating either one-sided quotient?
- If the left derivative is \(5\) and the right derivative is \(5\), what is the two-sided derivative, and why?
- If one one-sided quotient tends to \(+\infty\), is that a finite one-sided derivative under the definition here?
