Accumulation Derivatives with More Than One Formula
An accumulation function records the signed integral of a function from a fixed starting point to a variable endpoint. When the integrand is piecewise-defined, its formula changes across certain points. The derivative is still found locally: near a point where the integrand is continuous, the Fundamental Theorem of Calculus says that the accumulation derivative equals the integrand's value there.
The practical task is to identify which piece applies at the input. At a point where two pieces meet, continuity matters: if the integrand is continuous there, the same derivative rule applies at the joining point. At an endpoint of the domain, the corresponding statement uses a one-sided derivative.
Working One Piece at a Time
A piecewise-defined integrand has different formulas on specified intervals. To find \(F'(x)\), first determine the interval containing \(x\). If \(x\) lies in the interior of one of the intervals and that branch is continuous there, apply the accumulation derivative rule using that branch. There is no need to find a full formula for \(F(x)\) first.
At a joining point, check whether the pieces give the same value as the point is approached from either side, including the value assigned at the joining point. If they do, \(f\) is continuous there, so the derivative rule applies. The endpoint convention is slightly different: a function defined only on \([a,b]\) has no two-sided derivative at \(a\) or \(b\) as a function on an open interval. One can instead state its right derivative at \(a\) or its left derivative at \(b\).
Why the Rule Works Locally
The key is that differentiation examines what happens near one input \(c\). If \(c\) is strictly inside a piece, then sufficiently small changes in the input stay inside that piece. The other formulas do not affect the difference quotient at \(c\).
This argument also explains why the joining points require attention. An interior point of a single piece has a neighborhood governed by one continuous formula. At a joining point, values from both neighboring pieces can matter. If the integrand is continuous at that point, the continuity condition in the argument holds there as well.
Worked Examples
Worked Example: Two Polynomial Pieces Meeting Continuously
Let \(f\) be defined on \([0,4]\) by \(f(t)=t+1\) for \(0\leq t\leq2\), and \(f(t)=3t-3\) for \(2<t\leq4\). Define \(F(x)=\int_0^x f(t)\,dt\). Find the accumulation derivative on the domain, including at the joining point and the endpoints.
For \(0<x<2\), the first formula applies in a neighborhood of \(x\), and it is continuous there. Hence \(F'(x)=x+1\).
For \(2<x<4\), the second formula applies in a neighborhood of \(x\), so \(F'(x)=3x-3\).
At the joining point, the values from the two formulas agree: the first gives \(2+1=3\), and the second gives \(3(2)-3=6-3=3\). Thus \(f\) is continuous at \(2\), and \(F'(2)=f(2)=3\).
At the left endpoint, \(f(0)=0+1=1\), so the right derivative is \(F'_+(0)=1\). At the right endpoint, \(f(4)=3(4)-3=12-3=9\), so the left derivative is \(F'_-(4)=9\).
Check: The derivative on each open interval is exactly the branch of \(f\) defined there. At \(x=2\), both branch values equal \(3\), consistent with \(F'(2)=3\). The endpoint values also agree with their respective one-sided derivative rules.
Worked Example: Selecting the Branch at an Interior Input
Suppose \(f\) is defined on \([-1,3]\) by \(f(t)=t^2\) for \(-1\leq t\leq1\), and \(f(t)=2t-1\) for \(1<t\leq3\). Let \(G(x)=\int_0^x f(t)\,dt\). Find \(G'(x)\) on each interval and at the joining point.
When \(-1<x<1\), the applicable branch is \(t^2\). It is continuous, so \(G'(x)=x^2\).
When \(1<x<3\), use the branch \(2t-1\), giving \(G'(x)=2x-1\).
At \(x=1\), the first branch gives \(1^2=1\). The limiting value from the second branch is \(2(1)-1=2-1=1\). Since the assigned value at \(1\) is also \(1\), \(f\) is continuous there, and \(G'(1)=1\).
For completeness, the one-sided derivative at the left endpoint is \(G'_+(-1)=f(-1)=(-1)^2=1\). At the right endpoint, \(G'_-(3)=f(3)=2(3)-1=6-1=5\).
Check: The branch values coincide at \(1\), and the derivative there is \(1\). The interval formulas and endpoint values all use the same branch as the integrand at the relevant input.
Worked Example: A Trigonometric Piece and a Linear Piece
Define \(f\) on \([-\frac{\pi}{2},1]\) by \(f(t)=\cos t\) for \(-\frac{\pi}{2}\leq t<0\), and \(f(t)=1-t\) for \(0\leq t\leq1\). Let \(H(x)=\int_{-\pi/2}^x f(t)\,dt\). Find the derivative on both pieces, at the joining point, and at the endpoints.
For \(-\frac{\pi}{2}<x<0\), the cosine branch is continuous, so \(H'(x)=\cos x\). For \(0<x<1\), the linear branch is continuous, so \(H'(x)=1-x\).
At \(x=0\), the left-hand value approaches \(\cos 0=1\), and the right-hand branch gives \(1-0=1\). The assigned value is also \(1\). Therefore \(f\) is continuous at \(0\), and \(H'(0)=1\).
At the left endpoint, \(f(-\frac{\pi}{2})=\cos(-\frac{\pi}{2})=0\), so \(H'_+(-\frac{\pi}{2})=0\). At the right endpoint, \(f(1)=1-1=0\), so \(H'_-(1)=0\).
Check: The branch derivatives are \(\cos x\) and \(1-x\), matching \(f(x)\) on their respective intervals. Both pieces give \(1\) at the join, and both endpoint derivative values match the integrand at those endpoints.
A Common Branch-Selection Mistake
A common error is to use a familiar formula without checking the interval on which it is defined. In the second example, at \(x=\frac{3}{2}\), the input lies in \(1<x\leq3\), so the correct branch is \(2x-1\). The derivative is therefore \(G'(\frac{3}{2})=2(\frac{3}{2})-1=3-1=2\).
Using the first branch outside its interval would give the incorrect calculation \(G'(\frac{3}{2})=(\frac{3}{2})^2=\frac{9}{4}\). That formula defines \(f\) only when \(-1\leq x\leq1\), so it cannot be used at \(\frac{3}{2}\). The correction is to check the interval first, then substitute the input into the branch that applies.
Using the Rule Reliably
For a piecewise integrand, the calculation can be organized in three steps:
Determine which interval contains \(x\), paying attention to whether a boundary point is included in the left piece, the right piece, or both.
Where the selected branch is continuous, the accumulation derivative equals that branch evaluated at \(x\).
At a joining point, verify continuity before applying the rule. At an endpoint of the domain, state the derivative from within the domain as one-sided.
This method is useful whenever an accumulated quantity is built from a rate that follows different formulas over different intervals. A rate may change its expression while remaining continuous at a transition; in that case the local accumulation derivative is still given by the rate at the transition. For every input, however, the interval conditions and continuity hypotheses remain essential.
Check Your Understanding
Use the piecewise derivative rule and state any continuity or endpoint condition needed.
- If \(F(x)=\int_0^x f(t)\,dt\) and \(f(t)=t+2\) for \(0\leq t<1\), what is \(F'(x)\) for \(0<x<1\)?
- If the branch on \(1\leq t\leq3\) is \(f(t)=2t\), what is the value from that branch at \(t=1\)?
- What must be checked before applying the accumulation derivative rule at a point where two pieces meet?
- For an accumulation function defined on \([a,b]\), which one-sided derivative is relevant at \(a\)?
- In the second worked example, why is \(x^2\) not the correct branch for finding \(G'(\frac{3}{2})\)?
