Derivative: at a point

Section 10 Derivative: at a point

Objectives

Give formal limit definition of the derivative of a function at a point.
Understand the different quotient of a function in terms of rates of change and slopes of secants.
Understand the derivative of a function at a point both as an instantaneous rate of change, and as the slope of the tangent line to the graph at that point.

We will launch directly into an unmotivated definition of the derivative of a function $f$ at a point, and then motivate this definition retroactively in terms of important properties of $f$ and its graph.

Definition 10.1. Derivative at a point.

Suppose $a\in \R$ is an interior point or endpoint of the domain of $f\text{.}$ We say that $f$ is differentiable at $a$ if the limit

\begin{equation} \lim\limits_{h\to 0}\frac{f(h+a)-f(a)}{h},\tag{10.1} \end{equation}

or equivalently, the limit

\begin{equation} \lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}\text{,}\tag{10.2} \end{equation}

exists. When this is the case, we call this limit the derivative of $f$ at $a\text{,}$ denoted $f'(a)\text{:}$ i.e.,

\begin{equation} f'(a)= \lim\limits_{h\to 0}\frac{f(h+a)-f(a)}{h}=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}\text{.}\tag{10.3} \end{equation}

Remark 10.2. Derivative at a point.

Implicit in our definition is that the two limit expressions

\begin{align*} \lim\limits_{h\to 0}\frac{f(h+a)-f(a)}{h} \amp \amp \lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a} \end{align*}

are equivalent, in the sense that the one exists if and only if the other exists, in which case both are equal. This is seen to be true by invoking an appropriate affine substitution, which according to Remark 5.10, preserves the value of a limit.

In more detail, the $x\to a$ limit can be transformed to the $h\to 0$ limit by using the substitution $h=x-a\text{;}$ and conversely, the $h\to 0$ limit can be transformed to the $x\to a$ limit using the substitution $x=h+a\text{.}$

Remark 10.3. Limit definition of derivative.

Note that the limit

\begin{equation*} \lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h} \end{equation*}

always involves a denominator that approaches 0 as $h\to 0\text{.}$ This means we can never apply the quotient rule to such a limit directly; we will always have to do some algebra (or the like) to get to a place where we can potentially use limit rules.

Example 10.4. Derivative at point: elementary.

Let $f(x)=x^2\text{.}$

Show that $f$ is differentiable at $-2$ and compute $f'(-2)$ using the limit definition of the derivative.
More generally, fix any $a\in \R\text{.}$ Show that $f$ is differentiable at $a$ and compute $f'(a)$ using the limit definition of the derivative.

Solution.

We compute

\begin{align*} \lim\limits_{h\to 0 }\frac{f(-2+h)-f(-2)}{h} \amp = \lim\limits_{h\to 0}\frac{(-2+h)^2-4}{h} \\ \amp = \lim\limits_{h\to 0}\frac{4-4h+h^2-4}{h}\\ \amp = \lim\limits_{h\to 0}\frac{h(-4+h)}{h}\\ \amp = \lim\limits_{h\to 0}-4+h\\ \amp = -4 \amp \text{(poly. eval.)}\text{.} \end{align*}

Since the limit exists, we conclude that $f$ is differentiable at $-2\text{,}$ and that $f'(-2)=-4\text{.}$
The computation above is very similar for an arbitrary constant $a\in \R\text{:}$

\begin{align*} \lim\limits_{h\to 0 }\frac{f(a+h)-f(a)}{h} \amp = \lim\limits_{h\to 0}\frac{(a+h)^2-a^2}{h} \\ \amp = \lim\limits_{h\to 0}\frac{a^2+2ah+h^2-a^2}{h}\\ \amp = \lim\limits_{h\to 0}\frac{h(2a+h)}{h}\\ \amp = \lim\limits_{h\to 0}2a+h\\ \amp = 2a \amp \text{(poly. eval.)}\text{.} \end{align*}

We conclude that $f$ is differentiable at all $a\in \R\text{,}$ and that $f'(a)=2a\text{.}$

Definition 10.5. Difference quotient.

Let $f$ be a function. Given two distinct points $a\ne b$ in the domain of $f\text{,}$ the value

\begin{equation*} \frac{f(b)-f(a)}{b-a} \end{equation*}

is called a difference quotient of $f\text{.}$ We have two useful ways of understanding this quantity.

Average rate of change.

The difference quotient $\frac{f(b)-f(a)}{b-a}$ is called the average rate of change of $f$ between $a$ and $b\text{.}$ In this context we often use the $\Delta$ notation below to emphasize the relation to changing values of $f\text{:}$

\begin{equation*} \frac{f(b)-f(a)}{b-a}=\frac{\Delta f}{\Delta x}=\frac{\Delta y}{\Delta x}\text{,} \end{equation*}

where $y=f(x)\text{.}$
Slope of secant line.

Consider the points $P=(a,f(a))$ and $Q=(b,f(b))\text{,}$ which lie on the graph of $f\text{.}$ The unique line $L$ in $\R^2$ that passes through $P$ and $Q$ is called the secant line on $f$ between $P$ and $Q\text{.}$ This line can be defined, using point-slope form with respect to $P=(a,f(a))\text{,}$ by the equation

\begin{equation*} y-f(a)=\underset{\text{slope}}{\underbrace{\left (\frac{f(b)-f(a)}{b-a}\right)}}(x-a)\text{.} \end{equation*}

We recognize in this context that the difference quotient $\frac{f(b)-f(a)}{b-a}$ is the slope of the secant line $L\text{.}$

Definition 10.6. Derivative interpretations.

Let $f$ be differentiable at the point $a\text{.}$

Instantaneous rate of change.

We call the derivative value

\begin{equation*} f'(a)=\lim\limits_{h\to 0}\frac{f(h+a)-f(a)}{h}=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a} \end{equation*}

the instantaneous rate of change of $f$ at $a$ with respect to $x$, or more simply, the rate of change of $f$ at $a$ with respect to $a\text{.}$
Tangent line.

Let $P=(a,f(a))\text{.}$ The tangent line to the graph of $f$ at $P$ (or at $x=a$) is the line $L$ defined by the equation

\begin{equation*} y-f(a)=f'(a)(x-a)\text{.} \end{equation*}

In this context, we see that the derivative of $f$ at $a$ is the slope of the tangent line to the graph of $f$ at $P\text{.}$

Example 10.7. Tangent line at point.

Let $f(x)=\sqrt{5-x}\text{.}$

Sketch the graph of $f\text{.}$
Find an equation for the tangent line $L$ to the graph of $f$ at $x=1\text{.}$
Add $L$ to your sketch of the graph of $f\text{.}$

Solution.

Below you find the graph of $f(x)=\sqrt{5-x}\text{.}$ How do we come up with something like this by hand. The quick and dirty way is to remember what the graph of $g(x)=\sqrt{x}$ and realize that the graph of $f$ will have roughly the same shape, but perhaps shifted/scaled/reflected. A matter of a few plotted points (e.g., $(5,0)\text{,}$ $(4,1)\text{,}$ $(1,2)$), will then give us something like the following.

Figure 10.8. Graph of $f(x)=\sqrt{5-x}$
Let $L$ be the tangent line to the graph of $f$ at $x=1\text{.}$ By definition the slope of $L$ is $f'(1)\text{,}$ which we now compute:

\begin{align*} f'(1) \amp = \lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}\\ \amp = \lim\limits_{h\to 0}\frac{\sqrt{5-(1+h)}-\sqrt{5-1}}{h}\\ \amp = \lim\limits_{h\to 0}\frac{\sqrt{4-h}-2}{h}\\ \amp = \lim\limits_{h\to 0}\frac{\sqrt{4-h}-2}{h}\cdot\frac{\sqrt{4-h}+2}{\sqrt{4-h}+2}\\ \amp =\lim\limits_{h\to 0}\frac{4-h-4}{h(\sqrt{4-h}+2)} \\ \amp = \lim\limits_{h\to 0}\frac{-h}{h\sqrt{4-h}+2}\\ \amp = \lim\limits_{h\to 0}-\frac{1}{\sqrt{4-h}+2}\\ \amp = -\frac{1}{4} \amp \text{(cont.)}\text{,} \end{align*}

where the last step uses the fact that $g(h)=-\frac{1}{\sqrt{4-h}+2}$ is continuous at $0\text{.}$

We conclude that the slope of $L$ is $f'(1)=-\frac{1}{4}\text{.}$ We now use point-slope form to find an equation for $L\text{.}$ By definition the tangent line $L$ passes through the point $P=(1,f(1))=(1,2)$ on the graph of $f\text{.}$ With respect to this point we have the following point-slope equation for $L\text{:}$

\begin{equation*} y-2=-\frac{1}{4}(x-1)\text{.} \end{equation*}
We now add both the point $P$ and the tangent line $L$ to our graph of $f\text{.}$

Figure 10.9. Graph of $f(x)=\sqrt{5-x}\text{,}$ $P\text{,}$ and $L$

Prev Top Next