Differentiation rule for piecewise definition by interval

This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known.
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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:

$f(x):=\{\begin{array}{cc}{f}_1(x),& x<c\\ f_2(x),& c<x\\ v,& x=c\end{array}$

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 ${f_{1^{\prime }}}^{\prime }(c)$.
• $f$ is right differentiable at $c$ iff $v=f_2(c)$, and in this case, the right hand derivative equals ${f_{2^{\prime }}}^{\prime }(c)$.
• $f$ is differentiable at $c$ iff ($v=f_1(c)=f_2(c)$ and ${f_{1^{\prime }}}^{\prime }(c)={f_{2^{\prime }}}^{\prime }(c)$), and in this case, the derivative equals the equal values ${f_{1^{\prime }}}^{\prime }(c)$ and ${f_{2^{\prime }}}^{\prime }(c)$.

Piecewise definition of derivative

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

$f^{\prime }(x):=\{\begin{array}{cc}{f}_1\text{'}(x),& x<c\\ f_2\text{'}(x),& x>c\\ \end{array}$

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

$f^{\prime }(x):=\{\begin{array}{cc}{f}_1\text{'}(x),& x<c\\ f_2\text{'}(x),& x>c\\ u,& x=c\end{array}$

where $u={f_{1^{\prime }}}^{\prime }(c)={f_{2^{\prime }}}^{\prime }(c)$. 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:

$f(x):=\{\begin{array}{cc}{f}_1(x),& x<c\\ f_2(x),& c<x\le a_2\\ v,& x=c\end{array}$

Then, $f$ is $k$ times differentiable at $c$ if we have all these conditions: $f_1(c)=f_2(c)=v$, ${f_{1^{\prime }}}^{\prime }(c)={f_{2^{\prime }}}^{\prime }(c)$, $\dots$, $f_1^{(k)}(c)=f_2^{(k)}(c)$. 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</math> equals the equal values $f_1^{(k)}(c)=f_2^{(k)}(c)$.

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

$f^{(k)}(x):=\{\begin{array}{cc}{f}_1^{(k)}(x),& x<c\\ f_2^{(k)}(x),& x>c\\ u_k,& x=c\end{array}$

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 $\mathbb{R}$.