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Functions and single-variable calculus A course resource for Math 218

Ellen Goldstein, Aaron Greicius

Section 1.19 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 \(f\text{.}\)
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Definition 1.19.1. 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 \(f\).
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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.
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Example 1.19.2. 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.
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  2. Derive a formula for the derivative function \(f'\colon D\rightarrow \R\text{.}\)
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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 \(5\text{.}\)
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    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 \(5\text{.}\)
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  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*}
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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.
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Remark 1.19.3. 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.
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Remark 1.19.4. Failing to be differentiable.

By definition, a function \(f\) fails to be differentiable at an input \(a\) if the limit (1.83) 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 (1.83) 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.
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    This evident in the graph of \(f(x)=\sqrt{5-x}\text{.}\) (See Figure 1.18.8.) 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 \(x=5\text{.}\)
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  • Jump discontinuity \(\iff\) “corner point”.
    If the limit (1.83) 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 \(f\text{.}\)
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    The classic illustration of this phenomenon is the absolute value function \(f(x)=\abs{x}\text{.}\) As we show in Example 1.19.5 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 \(1\text{.}\)
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  • 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 1.11.2 as an example.
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Example 1.19.5. 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 \(f'\colon D\rightarrow \R\text{.}\)
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  2. Add a geometric argument about why \(f\) is not differentiable at \(0\) using Figure 1.19.6.
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Graph of absolute value
Figure 1.19.6. Graph of \(f(x)=\abs{x}\)
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Solution.
  1. We will show that \(f\) is differentiable at all points except \(a=0\text{.}\)
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    Since \(f(x)=\abs{x}=x\) for all positive \(x\text{,}\) it follows that
    \begin{equation*} f'(x)=\frac{d}{dx}(x)=1 \end{equation*}
    for all \(x\gt; 0\text{,}\) and hence that \(f'(a)=1\) for all \(a> 0\text{.}\) In particular, we see that \(f\) is differentiable at all positive points \(a> 0\text{.}\)
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    A similar argument shows that \(f'(a)=-1\) for all \(a< 0\text{,}\) and in particular that \(f\) is differentiable at all points \(a< 0\text{.}\)
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    To show that \(f\) is not differentiable at \(a=0\text{,}\) we show that the limit
    \begin{equation*} \lim\limits_{h\to 0}\frac{f(0+h)-f(0)}{h} \end{equation*}
    does not exist; and we do so by showing that the left and right limits are not equal. Indeed we have
    \begin{align*} \lim\limits_{h\to 0^+}\frac{f{0+h}-f(0)}{h}\amp =\lim\limits_{h\to 0^+}\frac{\abs{h}}{h}\\ \amp = \lim\limits_{h\to 0^+}\frac{h}{h} \amp (h> 0)\\ \amp = \lim\limits_{h\to 0^+}1=1 \end{align*}
    and
    \begin{align*} \lim\limits_{h\to 0^-}\frac{\abs{0+h}-f(0)}{h}\amp =\lim\limits_{h\to 0^-}\frac{-h}{h}\\ \amp = \lim\limits_{h\to 0^-}-1=-1\text{.} \end{align*}
    Since the left and right limits do not exist, we see that the limit \(\lim\limits_{h\to 0}\frac{f(0+h)-f(0)}{h}\) does not exist, and hence that \(f\) is not differentiable at \(a=0\text{.}\)
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  2. Geometrically, we see that \(f(x)=\abs{x}\) is not differentiable at \(a=0\text{,}\) as its graph has a corner point there. More precisely, there is no well-defined tangent line to the graph of \(f\) at \(a=0\text{.}\) Indeed, approaching \(a=0\) from the left, suggests that a tangent line at \(a=0\) would have equation \(y=-x\text{;}\) whereas approaching from the right, suggests the tangent line would have equation \(y=x\text{.}\)
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Example 1.19.7. Derivative function: \(\frac{1}{x}\).

Let \(f(x)=\frac{1}{x}\text{.}\) Compute the domain \(D\) of \(f'\) and derive a formula for \(f'\colon D\rightarrow \R\text{.}\)
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Solution.
We claim that \(f'(x)\) exists for all \(x\ne 0\) and thus that the domain of \(f'\) is
\begin{equation*} \{x\in \R\mid x\ne 0\}=(-\infty, 0)\cup (0,\infty)\text{.} \end{equation*}
(Note that we do not consider differentiability at \(x=0\) since \(f\) is not defined there.) Indeed, for any \(x\ne 0\text{,}\) we will show that \(f'(x)=-1/x^2\text{:}\)
\begin{align*} f'(x) \amp = \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ \amp = \lim\limits_{h\to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ \amp = \lim\limits_{h\to 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}\\ \amp = \lim\limits_{h\to 0}\frac{-h}{h\, x(x+h)}\\ \amp = \lim\limits_{h\to 0}-\frac{1}{x(x+h)}\\ \amp = -\frac{1}{x\cdot x} \amp \text{(alg. eval.)}\\ \amp = -\frac{1}{x^2}\text{.} \end{align*}
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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 (1.87).
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Definition 1.19.8. 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{1.86} \end{equation}
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  • 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{1.87} \end{equation}
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  • 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*}
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