# Differentiation rule for piecewise definition by interval

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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: $f(x) := \left\lbrace \begin{array}{rl} f_1(x), & x < c \\ f_2(x), & c < x \le a_2 \\v, & x = c \end{array}\right.$

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: $f(x) := \left\lbrace \begin{array}{rl} f_1(x), & x < c \\ f_2(x), & c < x \le a_2 \\v, & x = c \end{array}\right.$

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), $\dots$, $f_1^{(k)}(c) = f_2^{(k)}(c)$. 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$ 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 $\R$.