Differentiation rule for piecewise definition by interval

Statement

Everywhere version

Suppose $f_{1}$ and $f_{2}$ are functions of one variable, such that both of the functions are defined and differentiable everywhere. Consider a function $f$, defined as follows:

$f(x):=\left\lbrace {\begin{array}{rl}f_{1}(x),&x<c\\f_{2}(x),&c<x\leq a_{2}\\v,&x=c\end{array}}\right.$

Then, we have the following for continuity:

• The left hand limit of $f$ at $c$ equals $f_{1}(c)$.
• The right hand limit of $f$ at $c$ equals $f_{2}(c)$.
• $f$ is left continuous at $c$ iff $v=f_{1}(c)$.
• $f$ is right continuous at $c$ iff $v=f_{2}(c)$.
• $f$ is continuous at $c$ iff $f_{1}(c)=f_{2}(c)=v$.

We have the following for differentiability:

• $f$ is left differentiable at $c$ iff $v=f_{1}(c)$, and in this case, the left hand derivative equals $f_{1}'(c)$.
• $f$ is right differentiable at $c$ iff $v=f_{2}(c)$, and in this case, the right hand derivative equals $f_{2}'(c)$.
• $f$ is differentiable at $c$ iff ($v=f_{1}(c)=f_{2}(c)$ and $f_{1}'(c)=f_{2}'(c)$), and in this case, the derivative equals the equal values $f_{1}'(c)$ and $f_{2}'(c)$.

Piecewise definition of derivative

If the conditions for differentiability at $c$ are violated, we get the following piecewise definition for $f'$, which excludes the point $c$ from its domain:

$f'(x):=\left\lbrace {\begin{array}{rl}f_{1}'(x),&x<c\\f_{2}'(x),&x>c\\\end{array}}\right.$

If the conditions for differentiability at $c$ are satisfied, we get the following piecewise definition for $f'$, which includes the point $c$ in its domain:

$f'(x):=\left\lbrace {\begin{array}{rl}f_{1}'(x),&x<c\\f_{2}'(x),&x>c\\u,&x=c\end{array}}\right.$

where $u=f_{1}'(c)=f_{2}'(c)$. In particular, the value at $c$ can be included in either the left side or the right side definition.

Version for higher derivatives

Suppose $f_{1}$ and $f_{2}$ are functions of one variable, such that both of the functions are defined and $k$ times differentiable everywhere (and hence in particular the functions and their first $k-1$ derivatives are continuous), for some positive integer $k$. Consider the function:

$f(x):=\left\lbrace {\begin{array}{rl}f_{1}(x),&x<c\\f_{2}(x),&c<x\leq a_{2}\\v,&x=c\end{array}}\right.$

Then, $f$ is $k$ times differentiable at $c$ if we have all these conditions: $f_{1}(c)=f_{2}(c)=v$, $f_{1}'(c)=f_{2}'(c),<math>\dots$, $f_{1}^{(k)}(c)=f_{2}^{(k)}(c)$. In other words, the values should match, and the values of each of the derivatives up to the $k^{th}$ derivative should match. In that case, the $k^{th}$ derivative of $f$ at math>c</math> equals the equal values $f_{1}^{(k)}(c)=f_{2}^{(k)}(c)$.

The general piecewise definition of $f^{(k)}$ is, in this case:

$f^{(k)}(x):=\left\lbrace {\begin{array}{rl}f_{1}^{(k)}(x),&x<c\\f_{2}^{(k)}(x),&x>c\\u_{k},&x=c\end{array}}\right.$

where $u_{k}=f_{1}^{(k)}(c)=f_{2}^{(k)}(c)$.

Local generalization

The above holds with the following modification: we only require $f$ to be defined as $f_{1}$ on the immediate left of $c$ (i.e., on some interval of the form $(c-\delta ,c)$ for $\delta >0$ and as $f_{2}$ on the immediate right of $c$ (i.e., on some interval of the form $(c,c+\delta )$ for $\delta >0$). Further, we only require that $f_{1}$ and $f_{2}$ be defined and differentiable on open intervals containing $c$, not necessarily on all of $\mathbb {R}$.