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Section 11 Derivative: function

Whereas previously we focused on the notion of the derivative of a function \(f\) at a specific input, we now switch perspectives somewhat and consider the derivative operation as defining a function \(f'\) associated to \(f\text{.}\)

Definition 11.1. Derivative function.

Let \(f\) be a function, and let \(D\subseteq \R\) be the set of points where \(f\) is differentiable. The derivative operation defines a function
\begin{align*} f'\colon D \amp\rightarrow \R \\ x \amp \mapsto f'(x) \end{align*}
that is called the derivative of \(f\).
Note that the domain of the derivative function \(f'\) need not be the entire domain of \(f\text{:}\) indeed, by definition it consists precisely of the points where \(f\) is differentiable. The next example illustrates this.

Example 11.2. Derivative function: radical.

Let \(f\colon (-\infty, 5]\rightarrow \R\) be defined as \(f(x)=\sqrt{5-x}\text{.}\)
  1. Compute the domain \(D\) of \(f'\text{:}\) that is, find all points where \(f\) is differentiable.
  2. Derive a formula for the derivative function \(f'\colon D\rightarrow \R\text{.}\)
Solution.
  1. Take an element \(x\in (-\infty, 5]\) in the domain of \(f\) and compute:
    \begin{align*} \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h} \amp = \lim\limits_{h\to 0}\frac{\sqrt{5-(x+h)}-\sqrt{5-x}}{h}\\ \amp =\lim\limits_{h\to 0}\frac{\sqrt{5-(x+h)}-\sqrt{5-x}}{h}\cdot \frac{\sqrt{5-(x+h)}+\sqrt{5-x}}{\sqrt{5-(x+h)}+\sqrt{5-x}}\\ \amp = \lim\limits_{h\to 0}\frac{5-(x+h)-(5-x)}{h(\sqrt{5-(x+h)}+\sqrt{5-x})}\\ \amp = \lim\limits_{h\to 0}\frac{-h}{h(\sqrt{5-(x+h)}+\sqrt{5-x})}\\ \amp = \lim\limits_{h\to 0}-\frac{1}{\sqrt{5-(x+h)}+\sqrt{5-x}}\text{.} \end{align*}
    We now note that for all \(x\in (-\infty, 5)\text{,}\) this last limit expression evaluates to \(-\frac{1}{2\sqrt{5-x}}\text{,}\) after the application of simple limit rules. Thus \(f\) is differentiable for such elements. For \(x=5\text{,}\) however, the last limit becomes
    \begin{equation*} \lim\limits_{h\to 0}-\frac{1}{\sqrt{-h}} \end{equation*}
    which is a limit of type \(c/\infty\text{,}\) and hence infinite. In particular the limit doesn’t exist for \(x=5\text{,}\) showing that \(f\) is not differentiable at \(5\text{.}\)
    We conclude that the domain of \(f'\) is \(D=(-\infty, 5)\text{.}\) Note that \(D\) is strictly smaller than the domain of \(f\) in this case: indeed, it is every element of the domain of \(f\) with the exception of \(5\text{.}\)
  2. For \(x\ne 5\text{,}\) we can finish our limit computation above to compute \(f'(x)\text{:}\)
    \begin{align*} f'(x) \amp = \lim\limits_{h\to 0 }\frac{f(x+h)-f(x)}{h}\\ \amp = \lim\limits_{h\to 0}-\frac{1}{\sqrt{5-(x+h)}+\sqrt{5-x}}\\ \amp =-\frac{1}{\sqrt{\lim\limits_{h\to 0}5-(x+h)}-\sqrt{5-x}}\amp \text{(quot., root,sum)}\\ \amp = -\frac{1}{\sqrt{5-x}+\sqrt{5-x}} \amp \text{(poly. eval.)}\\ \amp = -\frac{1}{2\sqrt{5-x}}\text{.} \end{align*}
Using the slope interpretation of the derivative, we can begin to see how the derivative function \(f'\) implies information about the original function \(f\text{,}\) and conversely.

Remark 11.3. Slope interpretation.

Let \(D\) be the set of points where \(f\) is differentiable, and let \(f'\colon D\rightarrow \R\) be the derivative of \(f\text{.}\) Given any \(x\in D\text{,}\) we have the following equivalences:
\begin{align*} f'(x)> 0 \amp \iff \text{slope of tangent line to } f \text{ at } a \text{ is positive} \\ f'(x)< 0 \amp \iff \text{slope of tangent line to } f \text{ at } a \text{ is negative} \\ f'(x)=0 \amp \iff \text{tangent line to } f \text{ at } a \text{ is horizontal} \text{.} \end{align*}
We will elaborate further on these equivalences when we discuss the rate of change interpretation of the derivative in more detail.

Remark 11.4. Failing to be differentiable.

By definition, a function \(f\) fails to be differentiable at an input \(a\) if the limit (10.1) does not exist. The precise reason for that failure to exist is reflected in the graph of \(f\text{,}\) using the slope of tangent line interpretation of the derivative. Some examples and their graphical analogues:
  • Infinite limit \(\iff\) vertical slope.
    If the limit (10.1) is infinite at \(a\text{,}\) then the tangent line to the curve at \(a\) (if it exists) is vertical: i.e., its slope is undefined.
    This evident in the graph of \(f(x)=\sqrt{5-x}\text{.}\) (See Figure 10.8.) As \(x\to 5\text{,}\) \(f'(x)\to -\infty\text{;}\) equivalently, the slopes of the tangent lines to the graph of \(f\) at \(x\text{,}\) get arbitrarily large and negative. At \(x=5\) itself, the tangent line to the graph of \(f\) is in fact the vertical line \(x=5\text{.}\)
  • Jump discontinuity \(\iff\) “corner point”.
    If the limit (10.1) at \(a\) doesn’t exist because of a discrepancy between the left- and right-hand limits, both of which exist, then we typically see the slope of tangent lines at points \(x\) to the left of \(a\) approaching one limiting slope, while the slope of tangent lines of points \(x\) to the right of \(a\) will approach another value. Geometrically, such a point will look like a corner or kink in the graph of \(f\text{.}\)
    The classic illustration of this phenomenon is the absolute value function \(f(x)=\abs{x}\text{.}\) As we show in Example 11.5 this function is differentiable everywhere except at \(x=0\text{.}\) Graphically we see why this is the case in the corner nature of the point \(P=(0,0)\) on the graph: to the left the slope of the graph is \(-1\text{;}\) and to the right the slope is \(1\text{.}\)
  • Wild oscillation.
    If a function is wildly oscillating near an input \(a\) to such a degree that limit of of \(f\) doesn’t exist at \(a\text{,}\) then it turns out that the derivative also cannot exist at \(a\text{.}\) See the point at \(x=d\) for the function with graph in Figure 2.3 as an example.

Example 11.5. Derivative function: absolute value.

Let \(f(x)=\abs{x}\text{.}\)
  1. Show that the domain of \(f'\) is
    \begin{equation*} D=\R-\{0\}=\{x\in \R\colon x\ne 0\} \end{equation*}
    and compute a formula for \(f'\colon D\rightarrow \R\text{.}\)
  2. Add a geometric argument about why \(f\) is not differentiable at \(0\) using Figure 11.6.
Graph of absolute value
Figure 11.6. Graph of \(f(x)=\abs{x}\)
Solution.

Example 11.7. Derivative function: \(\frac{1}{x}\).

Let \(f(x)=\frac{1}{x}\text{.}\) Compute the domain \(D\) of \(f'\) and derive a formula for \(f'\colon D\rightarrow \R\text{.}\)
Solution.
We end this section by introducing an alternative notation for denoting the derivative of a function. The notation is called Leibniz notation, as it was favored in Gottlieb Leibniz’s development of calculus. Isaac Newton, on the other hand favored the prime notation \(f'\text{.}\) Leibniz notation comes in very handy when considering the derivative as an operation that we apply to functions; it also strongly evokes the interpretation of the interpretation of the derivative as a rate of change. It is very much less convenient as an alternative notation for \(f'(a)\text{,}\) as exhibited by (11.2).

Definition 11.8. Leibniz notation.

Let \(f\) be a function relating the output variable \(y\) as a function \(y=f(x)\) of the input variable \(x\text{,}\) and let \(D\subseteq \R\) be the set of points where \(f\) is differentiable. We introduce the following notation, called Leibniz notation, to denote various concepts related to \(f'\)
  • Derivative function.
    The derivative function \(f'\) is denoted as \(\displaystyle\frac{df}{dx}\) or \(\displaystyle\frac{dy}{dx}\text{.}\) In other words, we have have
    \begin{equation} f'=\frac{df}{dx}=\frac{dy}{dx}\text{.}\tag{11.1} \end{equation}
  • Derivative at a point.
    Given \(a\in D\text{,}\) the derivative \(f'(a)\) of \(f\) at \(a\) is denoted as \(\frac{df}{dx}\Bigr\vert_{x=a}\) or \(\frac{dy}{dx}\Bigr\vert_{x=a}\text{.}\) In other words, we have
    \begin{equation} f'(a)=\frac{df}{dx}\Bigr\vert_{x=a}=\frac{dy}{dx}\Bigr\vert_{x=a}\tag{11.2} \end{equation}
  • Derivative operation.
    We denote by \(\frac{d}{dx}\) the operation of taking the derivative of a function with respect to \(x\text{.}\) Thus \(\frac{d}{dx}(f)\) is the derivative of \(f\text{:}\) i.e.,
    \begin{equation*} \frac{d}{dx}(f)=f'\text{.} \end{equation*}