10.3 Definition of the Derivative.

Our definition of tangent to a curve will be based on the following definition:

10.18   Definition (Derivative.) Let $f$ be a real valued function such that $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$. Let $a\in\mbox{{\rm dom}}(f)$. We say that $f$ is differentiable at $a$ if $a$ is an interior point of $\mbox{{\rm dom}}(f)$ and the limit

$$\lim_{x\to a} {{f(x)-f(a)}\over {x-a}}$$ (10.19)

exists. In this case we denote the limit in (10.19) by $f'(a)$, and we call $f'(a)$ the derivative of $f$ at $a$.

The quantity $${ {{f(x)-f(a)}\over {x-a}}}$$ represents the slope of the line joining the points $\Big( a,f(a)\Big)$ and $\Big( x,f(x)\Big)$ on the graph of $f$. If $x$ and $a$ are different points in $\mbox{{\rm dom}}(f)$ then this quotient will be defined. If we choose a sequence of points $\{x_n\}$ converging to $a$, and if the slopes $${ \Big\{ {{f(x_n)-f(a)}\over {x_n-a}}\Big\}}$$ converge to a number $m$ which is independent of the sequence $\{x_n\}$, then it is reasonable to call $m$ (i.e., $f'(a)$) the slope of the tangent line to the graph of $f$ at $\Big( a,f(a)\Big)$.

10.20   Definition (Tangent to the graph of a function.) Let $f$ be a
real valued function with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$, and let $a\in\mbox{{\rm dom}}(f)$. If $f$ is differentiable at $a$ then we define the slope of the tangent to graph$(f)$ at the point $\Big( a,f(a)\Big)$ to be the number $f'(a)$, and we define the tangent to graph$(f)$ at $\Big( a,f(a)\Big)$ to be the line that passes through $\Big( a,f(a)\Big)$ with slope $f'(a)$.

Remark: This definition will need to be generalized later to apply to curves that are not graphs of functions. Also this definition does not allow vertical lines to be tangents, whereas on geometrical grounds, vertical tangents are quite reasonable.

10.21   Example. We will calculate the tangent to $\{y=x^3\}$ at a generic point $(a,a^3)$.

Let $f(x)=x^3$. Then for all $a\in\mbox{{\bf R}}$,

$$\begin{eqnarray*} \lim_{x\to a} {{f(x)-f(a)}\over {x-a}} &=& \lim_{x\to a} {{x^3... ...r {(x-a)}}\\ &=& \lim_{x\to a}(x^2+ax+a^2)=a^2+a^2+a^2=3a^2.\\ \end{eqnarray*}$$

Hence the tangent line to graph$(f)$ at $(a,a^3)$ is the line through $(a,a^3)$ with slope $3a^2$, and the equation of the tangent line is

$$y-a^3 = 3a^2(x-a)$$

or

$$y = a^3+3a^2x-3a^3=3a^2x-2a^3$$

or

$$y = a^2(3x-2a).$$

10.22   Example. We will now consider some of the examples given earlier.

If $f(x)=\vert x\vert$ then

$${{f(x)-f(0)}\over {x-0}}={{\vert x\vert}\over x}.$$

We saw in example 10.10 that $${\lim_{x\to 0}{{\vert x\vert}\over x}}$$ does not exist. Hence, the graph of $f$ at $(0,0)$ has no tangent.

If $g(x)=x^3$, then in the previous example we saw that the equation of the tangent to graph$(g)$ at $(0,0)$ is $y=0$; i.e., the $x$-axis is tangent to the curve. Note that in this case the tangent line crosses the curve at the point of tangency.

If $h(x)=x$ then for all $a\in\mbox{{\bf R}}$,

$$\lim_{x\to a} {{h(x)-h(a)}\over {x-a}} = \lim_{x\to a} {{x-a}\over {x-a}}=\lim_{x\to a}1=1.$$

The equation of the tangent line to graph$(h)$ at $(a,a)$ is

$$y=a+1(x-a)$$

or $y=x$.

Thus at each point on the curve the tangent line coincides with the curve.

Let $k(x)=(x^2)^{1/3}$. This is not the same as the function $l(x)=x^{2/3}$ since the domain of $l$ is $\mbox{{\bf R}}_{\geq 0}$ while the domain of $k$ is $\mbox{{\bf R}}$. (For all $x \in \mbox{{\bf R}}$ we have $x^2\in\mbox{{\bf R}}_{\geq 0}=\mbox{{\rm dom}}(g)$ where $g(x)=x^{1/3}$.)

I want to investigate $${ \lim_{x\to 0}{{k(x)-k(0)}\over {x-0}}=\lim_{x\to 0}{{k(x)}\over x}}$$. From the picture, I expect this graph to have an infinite slope at $(0,0)$, which means according to our definition that there is no tangent line at $(0,0)$. Let $\{x_n\}=\Big\{ \displaystyle { {1\over {n^3}}\Big\}}$. Then $\{x_n\} \to 0$, but

$${{k(x_n)}\over {x_n}}={{ \Big( {1\over {n^6}}\Big)^{1/3}}\ove... ...1\over {n^3}}\Big)}}={{ {1\over {n^2}}}\over { {1\over n^3}}}=n$$

so $${\lim \Big\{ {{k(x_n)}\over {x_n}}}\Big\}$$ does not exist and hence $${\lim_{x\to 0}{{k(x)}\over x}}$$ does not exist.

10.23   Example. Let $f(x)=\sqrt x$ for $x\in\mbox{{\bf R}}_{\geq 0}$. Let $a \in \mbox{${\mbox{{\bf R}}}^{+}$}$ and let $x\in\mbox{{\rm dom}}(f)\setminus\{a\}$. Then

$$\begin{eqnarray*} {{f(x)-f(a)}\over {x-a}}&=&{{\sqrt x-\sqrt a}\over {x-a}}={{\... ...\sqrt x-\sqrt a)(\sqrt x+\sqrt a)}}={1\over {\sqrt x +\sqrt a}}. \end{eqnarray*}$$

Hence

$$f'(a)=\lim_{x\to a} {{f(x)-f(a)}\over {x-a}}=\lim_{x\to a}{1... ...rt x+\sqrt a}}={1\over {\sqrt a+\sqrt a}}={1\over {2\sqrt a}};$$ (10.24)

i.e.,

$$f'(a)={1\over {2\sqrt a}} \mbox{ for all } a\in\mbox{${\mbox{{\bf R}}}^{+}$}.$$

In line (10.24) I used the fact that $${ \lim_{x\to a}\sqrt x=\sqrt a}$$, together with the sum and quotient rules for limits.

10.25   Exercise. A Let $f(x)=\displaystyle { {1\over x}}$. Sketch the graph of $f$. For what values of $x$ do you expect $f'(x)$ to be $-1$? For what values of $x$ do you expect $f'(x)$ to be positive? What do you expect to happen to $f'(x)$ when $x$ is a small positive number? What do you expect to happen to $f'(x)$ when $x$ is a small negative number?

Calculate $f'(a)$ for arbitrary $a\in\mbox{{\rm dom}}(f)$. Does your answer agree with your prediction?

10.26   Exercise. A Let $f(x)=\sin (x)$ for $-\pi<x<4\pi$. Sketch the graph of $f$. Use the same scale on the $x$-axis and the $y$-axis.

On what intervals do you expect $f'(x)$ to be positive? On what intervals do you expect $f'(x)$ to be negative? Calculate $f'(0)$.

On the basis of symmetry, what do you expect to be the values of $f'(\pi)$, $f'(2\pi )$ and $f'(3\pi )$? For what $x$ do you expect $f'(x)$ to be zero? On the basis of your guesses and your calculated value of $f'(0)$, draw a graph of $f'$, where $f'$ is the function that assigns $f'(x)$ to a generic number $x$ in $(-\pi,4\pi)$. On the basis of your graph, guess a formula for $f'(x)$.

(Optional) Prove that your guess is correct. (Some trigonometric identities will be needed.)

10.27   Exercise. A Calculate $f'(x)$ if $f(x)=\displaystyle { {x\over {x+1}}}$.

10.28   Exercise. A

a)

Find $f'(x)$ if $f(x)=x^2-2x$.

b)

Find the equations for all the tangent lines to graph$(f)$ that pass through the point $(0,-4)$. Make a sketch of graph$(f)$ and the tangent lines.

10.29   Exercise. Consider the function $f\colon (0,8)\to\mbox{{\bf R}}$ whose graph is shown below.

For what $x$ in $(0,8)$ does $f'(x)$ exist? Sketch the graphs of $f$ and $f'$ on the same set of axes.

The following definition which involves time and motion and particles is not a part of our official development and will not be used for proving any theorems.

10.30   Definition (Velocity.) Let a particle $\mbox{{\bf p}}$ move on a number line in such a way that its coordinate at time $t$ is $x(t)$, for all $t$ in some interval $J$. (Here time is thought of as being specified by a number.) If $t_0,t_1$ are points in $J$ with $t_0<t_1$, then the average velocity of $\mbox{{\bf p}}$ for the time interval $[t_0,t_1]$ is defined to be

$${{x(t_1)-x(t_0)}\over {t_1-t_0}}= {{ \mbox{ change in position } }\over {\mbox{ change in time } }}.$$

Note that $x(t_1)-x(t_0)$ is not necessarily the same as the distance moved in the time interval $[t_0,t_1]$. For example, if $x(t)=t(1-t)$ then $x(1)-x(0)=0$, but the distance moved by $\mbox{{\bf p}}$ in the time interval $[0,1]$ is $${ {1\over 2} }$$. (The particle moves from $0$ to $${ {1\over 4}}$$ at time $${t={1\over 2}}$$, and then back to $0$.)

The instantaneous velocity of $\mbox{{\bf p}}$ at a time $t_0\in J$ is defined to be

$$\lim_{t\to t_0} {{x(t)-x(t_0)}\over {t-t_0}}=x'(t_0)$$

provided this limit exists. (If the limit does not exist, then the instantaneous velocity of $\mbox{{\bf p}}$ at $t_0$ is not defined.) If we draw the graph of the function $x$; i.e., $\{ \Big( t,x(t)\Big)\colon t\in J\}$, then the velocity of $\mbox{{\bf p}}$ at time $t_0$ is by definition $x'(t_0)=$ slope of tangent to graph$(x)$ at $\Big( t_0,x(t_0)\Big)$.

In applications we will usually express velocity in units like $${ { {\rm miles}}\over { {\rm hour} }}$$. We will wait until we have developed some techniques for differentiation before we do any velocity problems.

The definition of velocity just given would have made no sense to Euclid or Aristotle. The Greek theory of proportion does not allow one to divide a length by a time, and Aristotle would no more divide a length by a time than he would add them. Question: Why is it that today in physics you are allowed to divide a length by a time, but you are not allowed to add a length to a time?

In Newton's calculus, the notion of instantaneous velocity or fluxion was taken as an undefined, intuitively understood concept, and the fluxions were calculated using methods similar to that used in the section 10.1.

The first ``rigorous'' definitions of limit of a function were given around 1820 by Bernard Bolzano (1781-1848) and Augustin Cauchy (1789-1857)[23, chapter 1]. The definition of limit of a function in terms of limits of sequences was given by Eduard Heine in 1872.