The derivative as the limit of a function

Summary

This lesson defines the derivative of a function at a point as a limit: the limit of the slopes of the secant lines, which is the slope of the tangent line, and the limit of the average velocities, which is the instantaneous velocity. It specifies at which points the definition makes sense, introduces one-sided derivatives, the tangent and normal lines, the derivative function and Leibniz notation, and proves that the derivative admits three equivalent formulations: the limit of the difference quotient, the approximation by a line with a negligible error, and the Carathéodory factorization. From the last one it follows that every differentiable function is continuous; the absolute value and the cube root show that the converse is false.

Learning objectives

Define the derivative of a function at a point as the limit of the difference quotient, and interpret it as the slope of the tangent line and as an instantaneous velocity.
Compute one-sided derivatives and use them to decide whether a function is differentiable at a point.
Prove the equivalence between the difference quotient, the linear approximation and the Carathéodory characterization.
Prove that every function differentiable at a point is continuous there, and recognize continuous functions that are not differentiable.

The tangent and the velocity

In the lesson on parabolas, a line is tangent to the curve at a point when it touches the curve only there without crossing it. That idea suffices for the conics, but not for an arbitrary curve. At the origin, the curve $y = x^3$ lies below the horizontal axis on the left and above it on the right: every line through the origin crosses it, and the lines $y = mx$ with $m < 0$ touch it only there, because $x^3 = mx$ requires $x = 0$ or $x^2 = m$ . No line satisfies the condition, and yet the horizontal axis is, to the eye, the line that best fits the curve near the origin. Another definition is therefore needed, one that depends not on the global shape of the curve but only on what happens near the point.

The idea is old: the tangent is the limiting position of the secant lines. Consider a function $f$ and a point $(a, f(a))$ on its graph. For $h \neq 0$ , the secant line through $(a, f(a))$ and through $(a + h, f(a + h))$ has slope

\frac{f(a + h) - f(a)}{h}

As $h$ approaches $0$ , the second point approaches the first, and if the slopes have a limit, the line through $(a, f(a))$ with that slope is the candidate for the tangent.

ExampleThe secants of a parabola

Let $f(x) = x^2$ and $a = 1$ . We compute the slopes of some secants through the point $(1, 1)$ and then their limit. The figure shows each secant and, at the end, the tangent.

Proof

$\frac{f(2) - f(1)}{1} = \frac{4 - 1}{1} = \resaltar{3}$
With $h = 1$ , the secant passes through $(1, 1)$ and $(2, 4)$ .
$\frac{f\left(\frac{3}{2}\right) - f(1)}{\frac{1}{2}} = \frac{\frac{9}{4} - 1}{\frac{1}{2}} = \resaltar{\frac{5}{2}}$
With $h = \frac{1}{2}$ , the second point is $\left(\frac{3}{2}, \frac{9}{4}\right)$ and the slope decreases.
$\frac{(1 + h)^2 - 1}{h} = \frac{2h + h^2}{h} = \resaltar{2 + h}$
In general, for every $h \neq 0$ , positive or negative, the slope simplifies; with $h = \frac{1}{4}$ it equals $\frac{9}{4}$ .
$\lim_{h \to 0} (2 + h) = 2, \qquad y = 1 + 2(x - 1) = \resaltar{2x - 1}$
As $h$ tends to $0$ , the slope tends to $2$ . The line through $(1, 1)$ with slope $2$ is the tangent; it is the same one given by the lemma on the tangent in the lesson on parabolas, with $x^2 = 4 \cdot \frac{1}{4} \, y$ .

The same limit appears when studying a motion. If $s(t)$ is the position, at time $t$ , of a point moving along a line, the quotient $\frac{s(t_0 + h) - s(t_0)}{h}$ is its average velocity between the times $t_0$ and $t_0 + h$ : the displacement divided by the elapsed time. The velocity at time $t_0$ , the one a speedometer would show, is the limit of the average velocities as $h$ tends to $0$ . For example, if $s(t) = t^2 - 4t$ , the average velocity between $1$ and $1 + h$ is $\frac{(1 + h)^2 - 4(1 + h) + 3}{h} = h - 2$ , which tends to $-2$ : at time $1$ , the point moves to the left at a rate of two units of length per unit of time. The two questions, that of the tangent and that of the velocity, lead to the same limit, and it is this limit that is defined below.

The derivative at a point

As in the lesson on the limit of a function, one must first specify where the limit makes sense. The quotient needs the value $f(a)$ , so $a$ must belong to the domain; and the limit as $x$ tends to $a$ requires $a$ to be an accumulation point of the domain.

DefinitionDerivative of a function at a point

Let $f$ be a real function defined on a set $A$ of real numbers, and $a$ an element of $A$ that is an accumulation point of $A$ . The difference quotient of $f$ at $a$ is the function $\frac{f(x) - f(a)}{x - a}$ , defined for those $x$ in $A$ that are different from $a$ . We say that $f$ is differentiable at $a$ if the difference quotient has a real limit $L$ as $x$ tends to $a$ ; that number is the derivative of $f$ at $a$ , and it is written $f'(a)$ :

f'(a) = L \leftrightarrow (\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{\frac{f(x) - f(a)}{x - a} - L} < \varepsilon\right)

The quotient is defined on $A \setminus \{a\}$ , and $a$ is an accumulation point of that set, because the definition of accumulation point looks only at the elements different from $a$ . By the uniqueness of the limit, the derivative, if it exists, is unique. At an isolated point of the domain the derivative is not defined, even though every function is continuous there. And if the quotient tends to $\infty$ or to $-\infty$ , the function is not differentiable at $a$ : the derivative is a real number. The definition is often written with the increment $h = x - a$ instead of the point $x$ , as in the example.

PropositionIncrement form of the derivative

Under the hypotheses of the definition, and considering $f(a + h)$ only where it is defined:

f'(a) = L \Leftrightarrow \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = L

This is the change of variable $x = a + h$ , which the translation and reflection proposition of the lesson on one-sided limits carried out for one side, now for both sides at once.

Proof

$Q(a + h) = \frac{f(a + h) - f(a)}{h}, \qquad h \neq 0 \rightarrow a + h \neq a$
Let $Q(x)$ be the difference quotient. The function $g(h) = a + h$ , for those $h \neq 0$ with $a + h$ in $A$ , tends to $a$ , does not take the value $a$ , and its values lie in the domain of $Q$ . By the theorem on composition of the lesson on the algebra of limits, if $Q$ tends to $L$ at $a$ , then $Q(a + h)$ also tends to $L$ at $0$ .
$Q(x) = Q\left(a + (x - a)\right), \qquad x \neq a \rightarrow \resaltar{x - a \neq 0}$
Conversely, the function $x - a$ , for those $x$ in $A$ that are different from $a$ , tends to $0$ and does not take the value $0$ ; composed with $h \mapsto Q(a + h)$ , it gives the quotient $Q(x)$ , which tends to the same limit.

The first derivatives are computed directly. If $f(x) = mx + b$ , the difference quotient equals $\frac{m(a + h) + b - (ma + b)}{h} = m$ for every $h \neq 0$ , and its limit is $m$ : the derivative of an affine function is its slope at every point, and its tangent line is the line itself. In particular, the derivative of a constant is $0$ and that of the identity $f(x) = x$ is $1$ . For $f(x) = x^2$ , the computation of the example, with $a$ instead of $1$ , gives $\frac{(a + h)^2 - a^2}{h} = 2a + h$ , and the derivative is $2a$ .

One-sided derivatives

At the endpoints of an interval, or at a point where the function changes formula, it is convenient to look at the difference quotient on each side separately.

DefinitionOne-sided derivatives

Let $a$ be an element of $A$ that is an accumulation point of $A \cap \ioo{a}{\infty}$ . We say that $f$ is differentiable from the right at $a$ if the difference quotient has a real right-hand limit at $a$ ; that limit is the right-hand derivative, $f'_+(a)$ :

f'_+(a) = L \leftrightarrow (\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(a < x < a + \delta \rightarrow \abs{\frac{f(x) - f(a)}{x - a} - L} < \varepsilon\right)

Likewise, if $a$ is an accumulation point of $A \cap \ioo{-\infty}{a}$ , the left-hand derivative, $f'_-(a)$ , is the left-hand limit of the difference quotient, with the condition $a - \delta < x < a$ .

PropositionDerivative and one-sided derivatives

Let $a$ be an element of $A$ .

If $a$ is an accumulation point of $A \cap \ioo{a}{\infty}$ and of $A \cap \ioo{-\infty}{a}$ , then $f$ is differentiable at $a$ if and only if the two one-sided derivatives exist and are equal; in that case, $f'(a) = f'_+(a) = f'_-(a)$ .
If $a$ is an accumulation point of the points of $A$ on one side only, then $f$ is differentiable at $a$ if and only if it is differentiable from that side, and the derivative is the one-sided derivative.

No new argument is needed: the one-sided derivatives are the one-sided limits of the difference quotient, and the lesson on one-sided limits already related them to the limit.

Proof

$\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = L \Leftrightarrow f'_+(a) = L = \resaltar{f'_-(a)}$
In the first case, the theorem on the limit and one-sided limits, applied to the difference quotient, gives the equivalence.
$\left(\forall x \in A\right)\left(0 < \abs{x - a} < r \rightarrow x > a\right) \rightarrow f'(a) = \resaltar{f'_+(a)}$
In the second, the same lesson observed that, when $a$ is an accumulation point of the points of the domain on one side only, the limit and the one-sided limit from that side coincide: there is an $r > 0$ such that all $x$ in $A$ with $0 < \abs{x - a} < r$ lie on that side.

For example, the absolute value has at $0$ the difference quotient $\frac{\abs{x}}{x}$ , which equals $1$ if $x > 0$ and $-1$ if $x < 0$ : its one-sided derivatives are $f'_+(0) = 1$ and $f'_-(0) = -1$ , which are different, and $\abs{x}$ is not differentiable at $0$ . On the other hand, if a function is defined on $\icc{0}{1}$ , its derivative at $0$ is its right-hand derivative, and at $1$ , its left-hand derivative: the definition already treats the endpoints of the domain as single sides. For example, $f(x) = x$ restricted to $\icc{0}{1}$ is differentiable at $0$ , with $f'(0) = 1$ .

Tangent and normal lines

With the derivative, the tangent ceases to be a figure recognized by eye and becomes a line that is computed.

DefinitionTangent line and normal line

Let $f$ be differentiable at $a$ . The tangent line to the graph of $f$ at the point $(a, f(a))$ is the line through it with slope $f'(a)$ :

y = f(a) + f'(a)(x - a)

The normal line at the same point is the perpendicular to the tangent through it: if $f'(a) \neq 0$ , it is the line $y = f(a) - \frac{1}{f'(a)}(x - a)$ , and if $f'(a) = 0$ , it is the vertical line $x = a$ .

The slope of the normal comes from the theorem on perpendicular lines of the lesson on the equation of a line: two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$ . Returning to the curve $y = x^3$ , its difference quotient at $0$ is $x^2$ , which tends to $0$ : the tangent at the origin is the horizontal axis, even though it crosses the curve. When $f$ is continuous at $a$ and the difference quotient tends to $\infty$ or to $-\infty$ , the secants approach the vertical; one then says that the graph has at $(a, f(a))$ a vertical tangent, the line $x = a$ , even though $f$ is not differentiable at $a$ .

Likewise, if $s(t)$ is the position of a point moving along a line, its velocity at time $t_0$ is $s'(t_0)$ , if it exists. Its sign indicates the direction of motion, and its absolute value, the speed. The lesson on rates of change will develop this reading.

The derivative function and Leibniz notation

Each point at which $f$ is differentiable contributes a number, its derivative there. With all of them a new function is formed.

DefinitionDerivative function

Let $f$ be a function defined on $A$ . The derivative function of $f$ is the function $f'$ that assigns to each point $x$ of $A$ at which $f$ is differentiable the number $f'(x)$ ; its domain is the set of those points. If $f$ is differentiable at every point of a set $S$ contained in $A$ , we say that it is differentiable on $S$ ; and if it is so on its whole domain, simply that it is differentiable.

For example, the derivative of $x^2$ is the function $2x$ , defined on all of $\mathbb{R}$ ; that of the absolute value equals $1$ on $\ioo{0}{\infty}$ and $-1$ on $\ioo{-\infty}{0}$ , and its domain does not contain $0$ . The notation $f'$ , with the prime, is due to Lagrange. The other usual notation is due to Leibniz: if $y = f(x)$ , the derivative is written $\frac{\mathrm{d}y}{\mathrm{d}x}$ , or $\frac{\mathrm{d}f}{\mathrm{d}x}$ , and the operation of differentiating with respect to $x$ is written $\frac{\mathrm{d}}{\mathrm{d}x}$ in front of the expression being differentiated:

\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) = 2x, \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(mx + b\right) = m

The notation recalls where the derivative comes from. If $\Delta x$ is the increment of the variable and $\Delta y = f(x + \Delta x) - f(x)$ the corresponding increment of the function, the difference quotient is $\frac{\Delta y}{\Delta x}$ , and the derivative is its limit:

\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}

It should be stressed that, for now, $\frac{\mathrm{d}y}{\mathrm{d}x}$ is a single symbol and not a quotient: $\mathrm{d}y$ and $\mathrm{d}x$ do not yet have a meaning on their own; they will in the lesson on higher-order derivatives and the differential. The notation makes explicit the variable with respect to which one differentiates: if position depends on time, velocity is written $s'(t)$ or $\frac{\mathrm{d}s}{\mathrm{d}t}$ .

Three characterizations of the derivative

The definition says that the difference quotient tends to $f'(a)$ . There are two other ways of saying the same thing, each useful for a different purpose: the second compares the function with its tangent line; the third avoids dividing.

TheoremCharacterizations of the derivative

Let $f$ be a function defined on $A$ , $a$ an element of $A$ that is an accumulation point of $A$ , and $L$ a real number. The following three statements are equivalent:

Difference quotient. $f$ is differentiable at $a$ and $f'(a) = L$ .
Linear approximation. $f(a + h) = f(a) + Lh + \operatorname{o}(h)$ as $h$ tends to $0$ ; that is, the remainder $R(h) = f(a + h) - f(a) - Lh$ satisfies $\lim_{h \to 0} \frac{R(h)}{h} = 0$ .
Carathéodory. There exists a function $\varphi \colon A \to \mathbb{R}$ , continuous at $a$ , with $\varphi(a) = L$ , such that $f(x) - f(a) = \varphi(x)(x - a)$ for every $x$ in $A$ .

The little $\operatorname{o}$ is the one from the lesson on equivalent infinitesimals and infinitely large functions, with $f(a + h)$ considered only where it is defined. The function $\varphi$ is the slope of the secant, completed at $a$ with the derivative. The figure shows, with $f(x) = \frac{x^2 + 1}{2}$ and $a = 1$ , the remainder between the curve and the tangent, a secant and, at the end, the graph of $\varphi$ , which in this case is the line $\varphi(x) = \frac{x + 1}{2}$ .

Proof

$\frac{R(h)}{h} = \frac{f(a + h) - f(a)}{h} - L \to 0 \Leftrightarrow \resaltar{\frac{f(a + h) - f(a)}{h} \to L}$
First and second: the quotient of the remainder by $h$ is the difference quotient minus $L$ . By the lemma on the limit as a distance that tends to zero, from the lesson on the algebra of limits, it tends to $0$ if and only if the difference quotient tends to $L$ , and by the increment form, this is equivalent to $f'(a) = L$ .
$\varphi(x) = \begin{cases} \frac{f(x) - f(a)}{x - a} & x \neq a \\ L & x = a \end{cases}, \qquad \lim_{x \to a} \varphi(x) = L = \resaltar{\varphi(a)}$
From the first to the third: define $\varphi$ as the difference quotient away from $a$ and as $L$ at $a$ . The equality holds at every $x \neq a$ by the definition, and at $x = a$ because both of its sides equal $0$ . Since $a$ is an accumulation point of $A$ and the limit of $\varphi$ at $a$ is $L = \varphi(a)$ , the proposition on continuity and limit of the lesson on limit and continuity gives the continuity of $\varphi$ at $a$ .
$x \neq a \rightarrow \frac{f(x) - f(a)}{x - a} = \varphi(x) \to \varphi(a) = \resaltar{L}$
From the third to the first: if $x \neq a$ , divide by $x - a$ , and the difference quotient is the restriction of $\varphi$ to $A \setminus \{a\}$ . Since $\varphi$ is continuous at $a$ , by the same proposition it tends to $\varphi(a) = L$ , and by the lemma on restriction, from the lesson on the algebra of limits, so does the quotient.

The function $\varphi$ of the third characterization, when it exists, is unique: away from $a$ it is the difference quotient, and at $a$ its value is the limit of that quotient. The second characterization explains in what sense the tangent is the line that best fits the curve. If $f(a + h)$ is replaced by $f(a) + mh$ , with a slope $m$ different from $f'(a)$ , the error is $(f'(a) - m)h + \operatorname{o}(h)$ , whose quotient by $h$ tends to $f'(a) - m \neq 0$ : it is of the same order as $h$ . Only with the slope $f'(a)$ is the error negligible compared with $h$ . The third characterization has a technical advantage: it is an equality between products, without divisions. The lessons on differentiation rules and on the chain rule will use it to avoid dividing by quantities that may vanish. The first consequence is immediate.

TheoremDifferentiable implies continuous

If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ . The Carathéodory factorization writes $f$ by means of operations between functions continuous at $a$ .

Proof

$f(x) = f(a) + \varphi(x)(x - a)$
By the theorem on characterizations, there exists $\varphi$ , continuous at $a$ , such that $f(x) = f(a) + \varphi(x)(x - a)$ for every $x$ in $A$ .
$\lim_{x \to a} f(x) = f(a) + \varphi(a) \cdot 0 = \resaltar{f(a)}$
The constant $f(a)$ and the polynomial $x - a$ are continuous at $a$ , and so is $\varphi$ . By the theorem on operations with continuous functions, from the lesson on limit and continuity, the sum of the constant and the product is continuous at $a$ .

The converse is false. Continuity requires $f(x) - f(a)$ to tend to $0$ ; differentiability requires more: that its quotient by $x - a$ have a limit, that is, that it tend to $0$ at least as fast as $x - a$ and without oscillating between different slopes.

PropositionContinuity without a derivative

The absolute value and the cube root are continuous at $0$ and not differentiable at $0$ : the first has different one-sided derivatives, and the difference quotient of the second tends to $\infty$ , so that its graph has a vertical tangent at the origin. Both are continuous by the corollary on known continuous functions, from the lesson on limit and continuity. The figure shows, for the absolute value, the two one-sided slopes and, for the cube root, ever steeper secants.

Proof

$f'_+(0) = \lim_{x \to 0^+} \frac{\abs{x}}{x} = 1 \neq -1 = \lim_{x \to 0^-} \frac{\abs{x}}{x} = \resaltar{f'_-(0)}$
The difference quotient of the absolute value at $0$ equals $1$ on the right and $-1$ on the left. The one-sided derivatives are different and, by the proposition on the derivative and one-sided derivatives, $\abs{x}$ is not differentiable at $0$ .
$\frac{\sqrt[3]{x} - 0}{x - 0} = \frac{1}{\left(\sqrt[3]{x}\right)^2} \to \resaltar{\infty}$
The difference quotient of the cube root at $0$ is the reciprocal of $\left(\sqrt[3]{x}\right)^2$ , which is positive if $x \neq 0$ and tends to $0$ by the continuity of the root. By the reciprocal lemma, from the lesson on infinite limits, the quotient tends to $\infty$ , and it has no real limit.

Both functions fail at a single point, but there are functions continuous on the whole line that are differentiable at no point; Weierstrass presented in 1872 the most famous example, whose construction is beyond this course.

Solved problems

The problems increase in difficulty: the first computes a derivative that ceases to exist at the edge of the domain; the second uses the tangent to discover a property of the hyperbola; the third shows that one-sided derivatives may fail to exist; the fourth, that a function can be differentiable at a single point; and the last combines the Carathéodory factorization with one-sided derivatives.

Solved problem 1The square root at the edge of its domain

Study the differentiability of $f(x) = \sqrt{x}$ , defined on $\ico{0}{\infty}$ , and find the tangent line at the point with abscissa $4$ .

Solution

$\frac{\sqrt{a + h} - \sqrt{a}}{h} = \frac{(a + h) - a}{\left(\sqrt{a + h} + \sqrt{a}\right) h} = \frac{1}{\sqrt{a + h} + \sqrt{a}}$
If $a > 0$ , the difference quotient is of the form $\frac{0}{0}$ ; multiply and divide by the conjugate expression.
$f'(a) = \resaltar{\frac{1}{2\sqrt{a}}}, \qquad y = 2 + \frac{1}{4}(x - 4) = \frac{x}{4} + 1$
By the continuity of the root, the denominator tends to $2\sqrt{a} \neq 0$ , and by the theorem on the limit of the reciprocal, from the lesson on the algebra of limits, the quotient tends to $\frac{1}{2\sqrt{a}}$ . At $a = 4$ , the derivative is $\frac{1}{4}$ and the tangent passes through $(4, 2)$ .
$\lim_{h \to 0^+} \frac{\sqrt{h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{\sqrt{h}} = \resaltar{\infty}$
At $a = 0$ there are points of the domain only on the right, and by the proposition on the derivative and one-sided derivatives, the derivative is the right-hand derivative. The quotient is $\frac{1}{\sqrt{h}}$ , and since $\sqrt{h}$ is positive and tends to $0$ , the reciprocal lemma gives $\infty$ .
$f \colon \ico{0}{\infty} \to \mathbb{R}, \qquad f' \colon \resaltar{\ioo{0}{\infty}} \to \mathbb{R}$
Therefore, $\sqrt{x}$ is differentiable on $\ioo{0}{\infty}$ and not at $0$ , where it is continuous and its graph has the vertical tangent $x = 0$ . The derivative function has a smaller domain than the function.

Solved problem 2The tangent to the hyperbola

Prove that the tangent line to $y = \frac{1}{x}$ at a point $\left(a, \frac{1}{a}\right)$ , with $a > 0$ , forms with the axes a triangle of area $2$ , whatever $a$ may be, and that the point of tangency is the midpoint of the line segment that the axes cut off on it.

Solution

$\frac{\frac{1}{a + h} - \frac{1}{a}}{h} = \frac{-h}{a(a + h) \, h} = -\frac{1}{a(a + h)} \to \resaltar{-\frac{1}{a^2}}$
The difference quotient simplifies on subtracting the fractions; by the theorem on the limit of the reciprocal, its limit is $-\frac{1}{a^2}$ .
$y = \frac{1}{a} - \frac{1}{a^2}(x - a) = \resaltar{\frac{2}{a} - \frac{x}{a^2}}$
The tangent passes through $\left(a, \frac{1}{a}\right)$ with that slope.
$\dato{c}{(2a, 0), \quad \left(0, \frac{2}{a}\right)}, \qquad \left(\frac{2a + 0}{2}, \frac{0 + \frac{2}{a}}{2}\right) = \resaltar{\left(a, \frac{1}{a}\right)}$
With $y = 0$ we obtain the intersection with the horizontal axis, and with $x = 0$ , the intersection with the vertical one. The midpoint between them is the point of tangency.
$\frac{1}{2} \cdot 2a \cdot \frac{2}{a} = \resaltar{2}$
The triangle has legs $2a$ and $\frac{2}{a}$ along the axes. Its area does not depend on $a$ : the figure shows two such triangles, of different shapes and the same area.

Solved problem 3Continuous and without one-sided derivatives

Let $f(x) = x \sin \frac{1}{x}$ and $g(x) = x^2 \sin \frac{1}{x}$ for $x \neq 0$ , and $f(0) = g(0) = 0$ . Prove that $f$ is continuous at $0$ and has at $0$ neither a derivative nor one-sided derivatives, and that $g$ is differentiable at $0$ , with $g'(0) = 0$ .

Solution

$\frac{f(x) - f(0)}{x - 0} = \frac{x \sin \frac{1}{x}}{x} = \resaltar{\sin \frac{1}{x}}$
By solved problem 3 of the lesson on limit and continuity, $f$ is continuous at $0$ . Its difference quotient at $0$ is $\sin \frac{1}{x}$ .
$\sin \frac{1}{x_n} = 1, \quad \sin \frac{1}{y_n} = -1, \qquad x_n = \frac{1}{\frac{\pi}{2} + 2n\pi}, \quad y_n = \frac{1}{\frac{3\pi}{2} + 2n\pi}$
Proposed problem 3 of the lesson on the limit proved that $\sin \frac{1}{x}$ has no limit at $0$ using two families of positive points, at which it equals $1$ and $-1$ . The same argument rules out the right-hand limit; with the opposite points, at which it equals $-1$ and $1$ , the left-hand one is ruled out.
$\frac{g(x) - g(0)}{x - 0} = x \sin \frac{1}{x} \to 0, \qquad g'(0) = \resaltar{0}$
On the other hand, the difference quotient of $g$ is $x \sin \frac{1}{x}$ , the product of a bounded function and another that tends to $0$ ; by the theorem of the lesson on the squeeze theorem, it tends to $0$ .
$\abs{g(x)} \leq x^2 = \resaltar{\operatorname{o}(x)}$
Both functions oscillate infinitely often near $0$ . The oscillation of $f$ is bounded by $\abs{x}$ , which is of the same order as the denominator $x - 0$ , and the quotient keeps oscillating; that of $g$ is bounded by $x^2$ , which is negligible compared with $x$ , and the quotient dies out.

Solved problem 4Differentiable at a single point

Let $f$ be the function that equals $x^2$ if $x$ is rational and $0$ if $x$ is irrational. Prove that $f$ is differentiable at $0$ , with $f'(0) = 0$ , and that it is not differentiable at any other point.

Solution

$x \in \mathbb{Q} \rightarrow x \, g(x) = x^2, \qquad x \notin \mathbb{Q} \rightarrow x \, g(x) = 0$
Let $g$ be the function of solved problem 5 of the lesson on limit and continuity, which equals $x$ on the rationals and $0$ on the irrationals. Then $f(x) = x \, g(x)$ for every $x$ , in both cases.
$\frac{f(x) - f(0)}{x - 0} = \frac{x \, g(x)}{x} = g(x) \to g(0) = 0, \qquad f'(0) = \resaltar{0}$
At $0$ , the difference quotient is $g(x)$ , which tends to $g(0) = 0$ because that problem proved that $g$ is continuous at $0$ .
$\lim_{x \to a} f(x) = f(a) \rightarrow \lim_{x \to a} \frac{f(x)}{x} = \frac{f(a)}{a} = \resaltar{g(a)}$
Let $a \neq 0$ . If $f$ were continuous at $a$ , so would be $g(x) = \frac{f(x)}{x}$ , by the quotient rule for operations with continuous functions, with a nonzero denominator at $a$ ; but the same problem proved that $g$ is discontinuous at $a$ . Therefore, $f$ is discontinuous at $a$ , and by the theorem that differentiable implies continuous, it is not differentiable at $a$ .

Solved problem 5The absolute value of a differentiable function

Let $f$ be a function defined on an open interval containing $a$ , differentiable at $a$ and with $f(a) = 0$ . Prove that $F = \abs{f}$ has at $a$ the one-sided derivatives $\abs{f'(a)}$ and $-\abs{f'(a)}$ , so that it is differentiable at $a$ if and only if $f'(a) = 0$ . Apply this to $\abs{\sin x}$ and to $\abs{x^3}$ at $0$ .

Solution

$\frac{F(x) - F(a)}{x - a} = \frac{\abs{f(x)} - \abs{f(a)}}{x - a} = \frac{\abs{\varphi(x)}\abs{x - a}}{x - a} = \begin{cases} \abs{\varphi(x)} & x > a \\ -\abs{\varphi(x)} & x < a \end{cases}$
By the Carathéodory characterization, $f(x) = \varphi(x)(x - a)$ , with $\varphi$ continuous at $a$ and $\varphi(a) = f'(a)$ . On taking absolute values, the factor $\abs{x - a}$ is $x - a$ on the right and $a - x$ on the left.
$F'_+(a) = \abs{f'(a)}, \qquad F'_-(a) = -\abs{f'(a)}, \qquad \abs{f'(a)} = -\abs{f'(a)} \Leftrightarrow \resaltar{f'(a) = 0}$
The function $\abs{\varphi}$ is continuous at $a$ , by the composition of continuous functions, and tends to $\abs{f'(a)}$ . By the proposition on the derivative and one-sided derivatives, $F$ is differentiable at $a$ if and only if the two one-sided derivatives coincide, that is, if $\abs{f'(a)} = -\abs{f'(a)}$ .
$\frac{\sin x}{x} \to 1 \neq 0, \qquad \frac{x^3}{x} = x^2 \to \resaltar{0}$
The sine satisfies $\sin 0 = 0$ , and its difference quotient at $0$ is $\frac{\sin x}{x}$ , which tends to $1$ by the fundamental trigonometric limit: $\abs{\sin x}$ has at $0$ one-sided derivatives $1$ and $-1$ , and is not differentiable. The cube has at $0$ the quotient $x^2$ , which tends to $0$ : $\abs{x^3}$ is differentiable at $0$ , with derivative $0$ .

Proposed problems

The problems increase in difficulty: the first reads a derivative in the geometry of the circle and the sphere; the second looks for normal lines through a given point; the third shows that a one-sided derivative can exist where the function jumps; the fourth compares the tangent given by the derivative with that of the circle; and the last studies a quotient that resembles the derivative without being one. Each one has a hint, which is best opened only after attempting the problem.

Proposed problem 1The area of the circle and the volume of the sphere

Let $A(r) = \pi r^2$ be the area of the circle of radius $r$ and $V(r) = \frac{4}{3}\pi r^3$ the volume of the sphere of radius $r$ . Compute, using the definition, $\frac{\mathrm{d}A}{\mathrm{d}r}$ and $\frac{\mathrm{d}V}{\mathrm{d}r}$ , and recognize each result.

Hint

Expand $(r + h)^2$ and $(r + h)^3$ and simplify the difference quotient before taking the limit. The increment of the area is the area of a ring of width $h$ .

Answer

\frac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r, \qquad \frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2

They are the circumference of the circle and the surface area of the sphere.

Proposed problem 2Normals through a point

Find all the points of the parabola $y = x^2$ whose normal line passes through the point $(0, 1)$ , and write those normals.

Hint

The normal at $(a, a^2)$ has slope $-\frac{1}{2a}$ if $a \neq 0$ ; require it to pass through $(0, 1)$ . Do not forget the case $a = 0$ , in which the derivative vanishes and the normal is vertical.

Answer

Three points: $(0, 0)$ , with normal $x = 0$ , and $\left(\pm\frac{\sqrt{2}}{2}, \frac{1}{2}\right)$ , with normals $y = 1 \mp \frac{x}{\sqrt{2}}$ .

Proposed problem 3The one-sided derivatives of the integer part

Prove that the integer part, $E(x) = \lfloor x \rfloor$ , is differentiable at every non-integer real number, with derivative $0$ , and that at each integer $n$ it has a right-hand derivative, $0$ , but no left-hand derivative. Does this contradict the theorem that differentiable implies continuous?

Hint

At an integer $n$ , on the right the function equals $n$ , and on the left, $n - 1$ : the difference quotient is $0$ on the right and $\frac{-1}{x - n}$ on the left. For the last question, compare the hypothesis of the theorem with what happens at $n$ .

Answer

E'(x) = 0 \quad (x \notin \mathbb{Z}), \qquad E'_+(n) = 0, \qquad \lim_{x \to n^-} \frac{E(x) - E(n)}{x - n} = \infty

There is no contradiction: the function is not differentiable at $n$ ; the right-hand derivative only gives continuity from the right (the theorem, applied to the restriction of $E$ to $\ico{n}{\infty}$ ), which holds.

Proposed problem 4The tangent to the circle

Let $f(x) = \sqrt{1 - x^2}$ , the upper semicircle with center $(0, 0)$ and radius $1$ . Compute $f'(a)$ for $-1 < a < 1$ , prove that the tangent is perpendicular to the radius joining the origin to $(a, f(a))$ , as in the lesson on the circle, and study what happens at $a = \pm 1$ .

Hint

Multiply and divide the difference quotient by $\sqrt{1 - (a + h)^2} + \sqrt{1 - a^2}$ . If $a \neq 0$ , the radius has slope $\frac{f(a)}{a}$ ; if $a = 0$ , it is vertical. At $a = 1$ there are points of the domain only on the left.

Answer

f'(a) = -\frac{a}{\sqrt{1 - a^2}}, \qquad -\frac{a}{\sqrt{1 - a^2}} \cdot \frac{\sqrt{1 - a^2}}{a} = -1 \quad (a \neq 0)

At $a = 0$ , the tangent is horizontal and the radius vertical. At $a = 1$ the quotient tends to $-\infty$ , and at $a = -1$ , to $\infty$ : $f$ is not differentiable at $\pm 1$ , and its graph has vertical tangents there.

Proposed problem 5The symmetric difference quotient

Let $f$ be defined on an open interval containing $a$ . Prove that, if $f$ is differentiable at $a$ , then $\lim_{h \to 0} \frac{f(a + h) - f(a - h)}{2h} = f'(a)$ . Prove with $f(x) = \abs{x}$ at $a = 0$ that this limit can exist without $f$ being differentiable.

Hint

Write the linear approximation for $h$ and for $-h$ and subtract them: the remainders give $\operatorname{o}(h)$ . For the absolute value, the numerator is $\abs{h} - \abs{-h}$ .

Answer

\frac{f(a + h) - f(a - h)}{2h} = f'(a) + \frac{R(h) - R(-h)}{2h} \to f'(a), \qquad \frac{\abs{h} - \abs{-h}}{2h} = 0

The absolute value has symmetric difference quotient $0$ at $0$ , and it is not differentiable there.

Worked solution

Solution

$f(a + h) = f(a) + f'(a)h + R(h), \qquad f(a - h) = f(a) - f'(a)h + R(-h)$
By the linear approximation, with $L = f'(a)$ , the remainder $R$ satisfies $\frac{R(h)}{h} \to 0$ . On replacing $h$ by $-h$ , by the theorem on composition, also $\frac{R(-h)}{-h} \to 0$ .
$\frac{f(a + h) - f(a - h)}{2h} = f'(a) + \frac{1}{2} \cdot \frac{R(h)}{h} + \frac{1}{2} \cdot \frac{R(-h)}{-h} \to \resaltar{f'(a)}$
Subtract the two equalities and divide by $2h$ ; by the theorem on the limit of a linear combination, the two quotients of the remainders give $0$ .
$\frac{\abs{h} - \abs{-h}}{2h} = \resaltar{0}, \qquad f'_+(0) = 1 \neq -1 = f'_-(0)$
For the absolute value at $0$ , the numerator is zero for every $h$ , and the symmetric difference quotient tends to $0$ ; however, the one-sided derivatives are $1$ and $-1$ . The symmetric difference quotient averages the two one-sided slopes and hides the corner.

The derivative as the limit of a function

Summary

Learning objectives

The tangent and the velocity

The derivative at a point

One-sided derivatives

Tangent and normal lines

The derivative function and Leibniz notation

Three characterizations of the derivative

Solved problems

Proposed problems

Your question about this lesson

Questions about this lesson