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Differential Calculus From Practice to Theory

Eugene Boman, Robert Rogers

Section 16.2 One Sided Derivatives

Since limits may be right-hand or left-hand and the derivative is defined as a limit, it follows that derivatives can be right– or left–hand as well. We will denote the right–hand and left—and derivatives at \(x=a\) with the notation: \(\rprime{f}(a)\text{,}\) and \(\lprime{f}(a)\text{,}\) respectively:
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Definition 16.2.1. One Sided Derivatives.

Given a function, \(f(x)\text{,}\) defined at a point \(a\text{:}\)
  1. \(f\) is said to have a right–hand derivative at \(x=a\) if the limit
    \begin{equation*} \rprime{f}(a)=\rlimit{h}{0}{\textstyle\frac{f(a+h)-f(a)}{h}} \end{equation*}
    exists.
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  2. \(f\) is said to have a left–hand derivative at \(x=a\) if the limit
    \begin{equation*} \lprime{f}(a)=\llimit{h}{0}{\textstyle\frac{f(a+h)-f(a)}{h}} \end{equation*}
    exists.
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From Theorem 16.1.13 and Definition 16.2.1 we see that the derivative of a function \(f\text{,}\) at a point \(x\) exists if and only if the right– and left–hand derivatives both exist, and are equal.
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At zero both the right– and left—and derivatives of the Absolute Value function \(f(x)=\abs{x}\) exist since
\begin{equation*} \rprime{f}(0)=\rlimit{h}{0}{\frac{\abs{0+h}-\abs{0}}{h}}=\rlimit{h}{0}{\frac{h}{h}}=1, \end{equation*}
and
\begin{equation*} \lprime{f}(0)=\llimit{h}{0}{\frac{\abs{0+h}-\abs{0}}{h}}=\llimit{h}{0}{\frac{-h}{h}}=-1. \end{equation*}
but \(f^\prime(0)\) does not exist because \(-1\neq1\text{.}\)
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Problem 16.2.3.

Notice that since the Heaviside function at \(x=0\) \(H(0)\text{,}\) is not defined, neither are \(\rprime{H}(0)\text{,}\) \(\lprime{H}(0)\text{,}\) or \(H^\prime(0)\text{.}\)
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(a)

Which of these derivatives would be defined if we were to arbitrarily define \(H(0)=1\text{?}\) Explain.
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(b)

Which of these derivatives would be defined if we were to arbitrarily define \(H(0)=-1\text{?}\) Explain.
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Problem 16.2.4.

Compute the right- and left-hand derivative, and the derivative itself, if they exist, of each of the following functions at the \(x=-2, -1, 0, 1,\) and \(2\text{.}\) If any of these derivatives don’t exist explain why not.
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Problem 16.2.5.

Use the limit definitions to compute the right– and left–hand derivative, of the function, \(f(x)\text{,}\) at \(x=-1\text{,}\) \(x=0\text{,}\) and \(x=2\text{.}\) At which of these points is \(f(x)\) differentiable?
\begin{equation*} f(x)= \begin{cases} 2\amp \text{ if }x\lt-1\\ -2(x+1)^2+2 \amp \text{ if }-1\le x \lt 0\\ x^4-4x \amp \text{ if }0\le x \lt 2\\ -14(x-3)^2+22 \amp \text{ if }2\le x \end{cases} \end{equation*}
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Problem 16.2.6.

(a)

Let \(H(x)\) be the Heaviside function and let \(a\) be a real number.
  1. Sketch \(H(x-a)\text{.}\)
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  2. Use Theorem 16.1.13 to show that \(H(x-a)\) is not differentiable at \(x=a\text{.}\)
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(b)

Let \(b\) also be a real number, distinct from \(a\text{,}\) and define
\begin{equation*} f(x) =H(x-a) + H(x-b). \end{equation*}
  1. Sketch \(f(x)\text{.}\)
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  2. Use Theorem 16.1.13 to show that is not differentiable at \(x=a\) or \(x=b\text{.}\)
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In Subsection 9.5.5 we found that the possible transition points come in three “flavors:”
  1. points where the derivative is zero,
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  2. points where the derivative is undefined, and,
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  3. if the function is defined on a closed interval, the endpoints of the interval.
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We remarked at the time that it seemed odd that the third condition does not involve the derivative while the other two do.
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With our new, deeper understanding of differentiation we can show that, in fact, there are really only two kinds of possible transition points, and both can be identified using the derivative. It turns out that the endpoints of a closed interval are not exceptional. They are really just points of non–differentiability.
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To see this consider any function, \(f(x)\) which is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\text{.}\) From Corollary 16.2.2 we know that \(f^\prime(b)\) only exists if \(\rprime{f}(b)\) and \(\lprime{f}(b)\) both exist.
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Aside: Comment.

But the right–hand limit at \(b\) doesn’t exist. To see this, observe that \(f\) is only defined on \([a,b]\text{.}\) Therefore when \(h>0\) the expression \(f(b+h)\) asks us to evaluate \(f\) at a point outside its domain, which is meaningless. Thus
\begin{equation*} \rprime{f}(b)=\rlimit{h}{0}{\frac{\CancelToRed{\text{meaningless}}{f(b+h)}-f(b)}{h}} \end{equation*}
does not exist. A similar argument shows that \(\lprime{f}(a)\) does not exist.
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Therefore \(f^\prime\) does not exist at the endpoints of a closed interval. As a result the possible transition points of \(f\) are only those places where the derivative is zero, or undefined, because the latter category includes the endpoints of closed intervals.
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