AP Calculus AB and BC glossary

Derivative of a Piecewise Function

Also called: Piecewise derivative, Differentiability of a piecewise function

The derivative of a piecewise function is the derivative of whichever piece the input falls in, which settles every point but the break. There the function must be continuous first, and the one sided derivatives from the left and right must agree on a finite value, or no derivative exists.

f(x)=\begin{cases} g(x), & x\le c\\ h(x), & x>c\end{cases} \quad\Longrightarrow\quad f'(x)=\begin{cases} g'(x), & x<c\\ h'(x), & x>c\end{cases}

Inside each open interval the function is just the rule that lives there, so if $f(x)=x^2$ for $x<1$ then $f'(x)=2x$ across that whole stretch. All the work is concentrated at the break point, and continuity is the first check there: where the pieces fail to meet, no arrangement of slopes rescues the derivative.

\lim_{h\to 0^-}\frac{f(c+h)-f(c)}{h} = \lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h}, \quad \text{both finite}

The second check compares those one sided difference quotients, and they have to agree on a finite value. Joining $\sqrt{x}$ on $x\ge 0$ to $-\sqrt{-x}$ on $x<0$ sends both of them to $+\infty$ , so they agree and yet no derivative exists, only a vertical tangent. Setting the two derivative formulas equal at the break point is a shortcut for that comparison, valid only when the one sided limits of $f'$ exist and are finite, which they do for the polynomial and rational pieces the exam uses. Joining $f(x)=x^2$ on $x\le 1$ to $ax+b$ on $x>1$ , continuity forces $a+b=1$ and equal slopes force $a=2$ , leaving $b=-1$ .

The mistake

Do the two slopes match at $x=c$ ? That settles only half of it. The derivative formulas set equal give one equation, but without the continuity equation the constants come out wrong, and a function with a jump is never differentiable there however neatly the slopes line up.

Appears in: Unit 2: Defining the Derivative