1.4 The derivative as best linear approximation

The derivative of f at a is the slope of the one straight line that hugs the graph better than any other near a. From that single idea come the rules of differentiation, the mean value theorem, and Taylor's theorem with an exact error bound.

In school the derivative was a slope, computed with rules. The rules worked, but where do they come from, and why does $f'(a)$ deserve the name “slope”? This chapter answers both with one idea: near $a$ , a differentiable function looks like a straight line, and $f'(a)$ is that line’s slope. That idea is also how derivatives work in many variables (Unit 5), where “slope” is replaced by a linear map from chapter 2.3.

Before you start

Chapter 1.3: limits of functions with $\varepsilon$ and $\delta$ , continuity, and the extreme value theorem.

Quick checkWhat does

\lim_{x \to a} g(x) = L

ignore?

Quick checkA continuous function on

[a, b]

always:

The picture: zoom until it is straight

Zoom in on $x^2$ at any point: by the time you have zoomed a hundredfold, the curve and the dashed tangent line are indistinguishable. The box below the picture shows why in numbers: the tangent’s error, divided by the step $h$ , shrinks towards 0. Two things to do:

Zoom on a smooth curve. Pick $x^2$ , then $\sin x$ , then $x^3 - x$ . Slide the point $a$ to a few places and zoom each time. Watch the number “error divided by $h$ ” fall as you zoom. That falling number is what the definition below formalises.
Zoom on the corner. Pick $|x|$ . At its corner, every zoom shows the same V. No line fits it, at any scale.

“Differentiable at $a$ ” means exactly that zooming in at $a$ produces a straight line.

Where this came from· Fermat's tangents, Newton's fluxions, Cauchy's limits

Pierre de Fermat wanted the highest point of a curve and the tangent at a point. In the 1630s he compared the value of a quantity at $x$ and at $x + e$ , divided by $e$ , and then set $e$ to zero. That is a difference quotient used to find a maximum, and his idea survives as the theorem below that the derivative vanishes at an interior maximum. Isaac Newton (his fluxions date from the mid-1660s) wanted velocities of moving bodies; Gottfried Leibniz, who published first, in 1684, wanted tangents and areas, and gave us the notation $\frac{dy}{dx}$ . Both worked with “infinitely small” increments that nobody could define exactly. Cauchy’s 1823 lectures at the École polytechnique defined the derivative as the limit of the difference quotient, the definition used in this chapter.

Definitions

Definition B

Derivative

Let $f$ be defined on an interval containing $a$ . The derivative of $f$ at $a$ is

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

if this limit exists. The fraction is the difference quotient: the slope of the chord from $(a, f(a))$ to $(a + h, f(a + h))$ . Here $a$ is fixed and $h$ is the variable, so this is a limit of a function of $h$ as in 1.3. The step $h$ is never 0 inside the limit, which is why the fraction makes sense.

In one line $f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ , when the limit exists: the slope of the best straight-line fit to f at a.

Why it existsIt turns 'rate of change' and 'tangent slope' into one exact number, and it is the slope of the unique line that approximates f near a to better than first order.

Definition B

Differentiable

$f$ is differentiable at $a$ if $f'(a)$ exists. Equivalently, there is a number $m$ such that

f(a + h) = f(a) + m\,h + E(h), \qquad \text{where } \frac{E(h)}{h} \to 0 \text{ as } h \to 0,

and then $m = f'(a)$ . In words: $f$ near $a$ is a straight line of slope $m$ plus an error that is small even compared with $h$ .

In one line $f$ is differentiable at $a$ if $f'(a)$ exists; equivalently, $f(a + h) = f(a) + f'(a)h + E(h)$ with $E(h)/h \to 0$ .

Why it existsThe second form says f is a line plus an error much smaller than h, which is the zoom picture and the definition that generalises to many variables.

Why· Why the two forms say the same thing

Divide $f(a + h) = f(a) + mh + E(h)$ by $h$ : $\frac{f(a+h) - f(a)}{h} = m + \frac{E(h)}{h}$ . The left side tends to $m$ exactly when $\frac{E(h)}h \to 0$ . The second form is the zoom picture: the error of the line $f(a) + mh$ shrinks faster than $h$ , so at fine scales it vanishes from view. It is also the form that works in $\mathbb R^n$ , where you cannot divide by a vector $h$ but you can still ask that the error be small compared with $|h|$ (Apostol 8.11, chapter 5.3).

Worked example

Find the derivative of

f(x) = x^2

a

from the definition, in both forms.

Solution.

Difference quotient. $\frac{(a + h)^2 - a^2}{h} = \frac{2ah + h^2}{h} = 2a + h$ for $h \ne 0$ . Expand and cancel. Cancelling $h$ is allowed because $h \ne 0$ inside a limit.
Take the limit. $2a + h \to 2a$ as $h \to 0$ . So $f'(a) = 2a$ . $2a + h$ is within $\varepsilon$ of $2a$ when $|h| < \varepsilon$ : take $\delta = \varepsilon$ .
Linear-approximation form. $(a + h)^2 = a^2 + 2a\,h + h^2$ , so $E(h) = h^2$ , and $\frac{E(h)}{h} = h \to 0$ . The same answer, read as ‘line plus small error’.

Answer. $f'(a) = 2a$ . Near $a = 1$ , $x^2 \approx 1 + 2(x - 1)$ , with error $(x - 1)^2$ .

Your answerFor

f(x) = x^2

a = 3

, compute the difference quotient

\frac{f(3 + h) - f(3)}{h}

with

h = 0.1

Worked example

Find the derivative of

f(x) = \sqrt x

a > 0

from the definition.

Solution.

Write the difference quotient. $\frac{\sqrt{a + h} - \sqrt a}{h}$ . The definition. Setting $h = 0$ gives $\frac00$ , so we must simplify before taking the limit.
Multiply top and bottom by the conjugate. $\frac{(\sqrt{a + h} - \sqrt a)(\sqrt{a + h} + \sqrt a)}{h\,(\sqrt{a + h} + \sqrt a)} = \frac{(a + h) - a}{h\,(\sqrt{a + h} + \sqrt a)}$ . The same rationalising trick you used in JEE. It turns the difference of square roots into $h$ , which can cancel.
Cancel $h$ . $= \frac1{\sqrt{a + h} + \sqrt a}$ for $h \ne 0$ . Allowed, because inside a limit $h \ne 0$ .
Take the limit. As $h \to 0$ , $\sqrt{a + h} \to \sqrt a$ (continuity of $\sqrt x$ , 1.3), so the quotient tends to $\frac1{2\sqrt a}$ . The denominator tends to $2\sqrt a$ , which is not 0 because $a > 0$ , so the quotient law applies.

Answer. $f'(a) = \frac1{2\sqrt a}$ . $\blacksquare$ At $a = 4$ this is $\frac14$ , the slope used in the $\sqrt{4.1}$ estimate below.

Your answerFrom the definition, the difference quotient of

f(x) = x^3

a = 2

\frac{(2 + h)^3 - 8}{h} = 12 + 6h + h^2

. What is

f'(2)

Quick checkWhich error

E(h)

, in

f(a + h) = f(a) + mh + E(h)

, is small enough for

m

to be the derivative?

Worked example

Show

f(x) = |x|

is not differentiable at 0.

Solution.

Difference quotient. $\frac{|h| - 0}{h}$ equals $1$ for $h > 0$ and $-1$ for $h < 0$ . $|h| = h$ or $-h$ depending on the sign.
Use sequences (1.3). Along $h_n = \frac1n$ the quotient is 1; along $h_n = -\frac1n$ it is $-1$ . A limit would have to equal both. If a limit exists, every sequence $h_n \to 0$ gives the same value (the sequential criterion of 1.3, in its limit form).

Answer. No limit, so $|x|$ is not differentiable at 0, although it is continuous there. The zoom picture shows the V that never straightens.

Theorem B Abbott Thm 5.2.3

Differentiable implies continuous

If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ . (The converse fails: $|x|$ .)

In one lineIf $f$ is differentiable at $a$ , it is continuous at $a$ .

Why it is truef(a + h) − f(a) = h · (difference quotient), and h → 0 while the quotient stays near f'(a).

ProofAbbott 5.2.3

The idea: Write f(a + h) − f(a) as h times the difference quotient. The first factor tends to 0, the second to f'(a), so the product tends to 0.

For

h \ne 0

f(a + h) - f(a) = h \cdot \dfrac{f(a + h) - f(a)}{h}

Multiply and divide by

h \ne 0

h \to 0

the right side tends to

0 \cdot f'(a) = 0

Product of limits (the limit laws of 1.2 carry over to function limits through the sequential criterion of 1.3).

f(a + h) \to f(a)

f

is continuous at

a

\blacksquare

That is the definition of continuity at

a

, with

x = a + h

Quick checkWhich of these is true?

The rules, from the definition

Theorem B Spivak ch. 10

Chain rule

If $g$ is differentiable at $a$ and $f$ is differentiable at $g(a)$ , then $f \circ g$ is differentiable at $a$ with $(f \circ g)'(a) = f'(g(a))\, g'(a)$ .

In one lineIf $g$ is differentiable at $a$ and $f$ at $g(a)$ , then $(f \circ g)'(a) = f'(g(a))\,g'(a)$ .

Why it is trueApproximating each map by a line, composing two lines multiplies their slopes.

Why· The idea in one line, and why the naive proof fails

Linear approximations compose: if $g$ multiplies small changes by $g'(a)$ and $f$ multiplies small changes by $f'(g(a))$ , doing one after the other multiplies by the product. The tempting proof writes $\frac{f(g(a+h)) - f(g(a))}{h} = \frac{f(g(a+h)) - f(g(a))}{g(a+h) - g(a)} \cdot \frac{g(a+h) - g(a)}{h}$ , but $g(a + h) - g(a)$ can be 0 for small $h$ (for example if $g$ is constant), and then the first fraction is undefined. The proof below avoids dividing by it.

ProofSpivak ch. 10, Abbott 5.2.5

The idea: Define a function φ(k) that equals the difference quotient of f at g(a) when k ≠ 0 and equals f'(g(a)) when k = 0. Then f(g(a) + k) − f(g(a)) = φ(k)·k for EVERY k, including 0, and φ is continuous at 0. Substitute k = g(a + h) − g(a).

Let

b = g(a)

, and define

\varphi(k) = \dfrac{f(b + k) - f(b)}{k}

for

k \ne 0

, and

\varphi(0) = f'(b)

Abbreviate.

b = g(a)

\varphi

is continuous at 0.

Differentiability of

f

b

says exactly that the difference quotient tends to

f'(b) = \varphi(0)

.This patch at

k = 0

is the whole trick.

For every

k

f(b + k) - f(b) = \varphi(k)\, k

For

k \ne 0

multiply the definition of

\varphi

k

; for

k = 0

both sides are 0.

\dfrac{f(g(a + h)) - f(g(a))}{h} = \varphi\big(g(a + h) - g(a)\big) \cdot \dfrac{g(a + h) - g(a)}{h}

Substitute

k = g(a + h) - g(a)

, so

b + k = g(a + h)

, and divide by

h \ne 0

h \to 0

the right side tends to

\varphi(0) \cdot g'(a) = f'(g(a))\,g'(a)

\blacksquare

g

is continuous at

a

(differentiable implies continuous), so

g(a + h) - g(a) \to 0

;

\varphi

is continuous at 0; the second factor tends to

g'(a)

Worked example

See the patch

\varphi

at work:

f(u) = u^2

composed with

g(x) = x^3

a = 1

Solution.

Set the pieces. $b = g(1) = 1$ , $f'(b) = 2$ , $g'(1) = 3$ . The chain rule predicts $(f \circ g)'(1) = f'(1)\,g'(1) = 2 \times 3 = 6$ .
Build $\varphi$ . For $k \ne 0$ : $\varphi(k) = \frac{(1 + k)^2 - 1}{k} = \frac{2k + k^2}{k} = 2 + k$ . At $k = 0$ define $\varphi(0) = f'(1) = 2$ . The formula $2 + k$ already gives 2 at $k = 0$ , so the patch costs nothing here. For a general $f$ the quotient is undefined at $k = 0$ and the patch is needed.
Check the key identity. $f(1 + k) - f(1) = (1 + k)^2 - 1 = (2 + k)\,k = \varphi(k)\,k$ , true for every $k$ including $k = 0$ . This is the line that avoids dividing by $g(a + h) - g(a)$ .
Substitute $k = g(1 + h) - g(1) = (1 + h)^3 - 1 = 3h + 3h^2 + h^3$ . Then $\frac{f(g(1 + h)) - f(g(1))}{h} = \varphi(k)\cdot\frac{k}{h} = (2 + k)(3 + 3h + h^2)$ . Here $\frac kh = \frac{g(1 + h) - g(1)}{h}$ , the difference quotient of $g$ .
Let $h \to 0$ . Then $k \to 0$ , so $2 + k \to 2$ , and $3 + 3h + h^2 \to 3$ . The product tends to $6$ . Matches the prediction. Direct check: $f(g(x)) = x^6$ has derivative $6x^5 = 6$ at 1.

Answer. $(f \circ g)'(1) = 6 = f'(g(1))\,g'(1)$ . $\blacksquare$

Proof workshopPick the reason for each line of the chain rule proof

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Let $b = g(a)$ and define $\varphi(k) = \frac{f(b + k) - f(b)}{k}$ for $k \ne 0$ , and $\varphi(0) = f'(b)$ .
$\varphi$ is continuous at 0.
For every $k$ , $f(b + k) - f(b) = \varphi(k)\,k$ .
Put $k = g(a + h) - g(a)$ and divide by $h \ne 0$ : $\frac{f(g(a+h)) - f(g(a))}{h} = \varphi\big(g(a+h) - g(a)\big)\,\frac{g(a+h) - g(a)}{h}$ .
As $h \to 0$ , $g(a + h) - g(a) \to 0$ .
So the right side tends to $\varphi(0)\,g'(a) = f'(g(a))\,g'(a)$ .

Quick checkA student proves the chain rule by writing

\frac{f(g(a+h)) - f(g(a))}{h} = \frac{f(g(a+h)) - f(g(a))}{g(a+h) - g(a)}\cdot\frac{g(a+h) - g(a)}{h}

. What is wrong?

The sum, product and quotient rules are proved the same way. The product rule uses the add-and-subtract trick of 1.2: $f(a+h)g(a+h) - f(a)g(a) = f(a+h)\big(g(a+h) - g(a)\big) + g(a)\big(f(a+h) - f(a)\big)$ .

Worked example Abbott Thm 5.2.4

Prove the product rule:

(fg)'(a) = f'(a)\,g(a) + f(a)\,g'(a)

Solution.

Write the difference quotient of $fg$ . $\frac{f(a + h)g(a + h) - f(a)g(a)}{h}$ . The definition of $(fg)'(a)$ .
Add and subtract $f(a + h)g(a)$ in the numerator. The numerator becomes $f(a + h)\big(g(a + h) - g(a)\big) + g(a)\big(f(a + h) - f(a)\big)$ . The aim is to produce difference quotients of $f$ and of $g$ separately. This is the same trick as the product law for sequences in 1.2: change one factor at a time.
Split the quotient. $f(a + h)\cdot\frac{g(a + h) - g(a)}{h} + g(a)\cdot\frac{f(a + h) - f(a)}{h}$ . Divide each part by $h$ .
Take limits. The two quotients tend to $g'(a)$ and $f'(a)$ . Also $f(a + h) \to f(a)$ . $f$ is differentiable at $a$ , hence continuous there, by the theorem above.
Combine. The sum tends to $f(a)\,g'(a) + g(a)\,f'(a)$ . Sum and product laws for limits.

Answer. $(fg)'(a) = f'(a)\,g(a) + f(a)\,g'(a)$ . $\blacksquare$

The mean value theorem

Theorem B Abbott Thm 5.2.6 (Fermat)

Interior extremum theorem

If $f$ is differentiable at an interior point $c$ of an interval and $f(c) \ge f(x)$ for all $x$ in the interval (or $\le$ ), then $f'(c) = 0$ .

In one lineIf f has a maximum or minimum at an interior point c and is differentiable there, then f'(c) = 0.

Why it is trueAt a maximum, chords to the right slope down and chords to the left slope up, so the limit is squeezed to 0.

ProofAbbott 5.2.6

The idea: At a maximum, f(c + h) − f(c) ≤ 0. Dividing by h > 0 gives quotients ≤ 0; dividing by h < 0 gives quotients ≥ 0. The common limit f'(c) is both ≤ 0 and ≥ 0.

For small

h

f(c + h) - f(c) \le 0

c

is a maximum, and

c + h

is in the interval for small

|h|

because

c

is interior.

For

h > 0

the difference quotient is

\le 0

; for

h < 0

it is

\ge 0

Dividing by a positive

h

keeps the sign; by a negative

h

flips it.

f'(c) \le 0

and

f'(c) \ge 0

, hence

f'(c) = 0

\blacksquare

Limits preserve non-strict inequalities (take limits along

h_n = \pm\frac1n

Theorem B Abbott Thm 5.3.2

Mean value theorem

Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$ . Then there is $c \in (a, b)$ with

f'(c) = \frac{f(b) - f(a)}{b - a}.

The special case $f(a) = f(b)$ , where $f'(c) = 0$ , is Rolle’s theorem.

In one lineIf f is continuous on [a, b] and differentiable on (a, b), some c in (a, b) has f'(c) = (f(b) − f(a))/(b − a).

Why it is trueTilt the graph so the chord is level; the tilted function has a maximum or minimum inside, where its derivative is 0.

Why· The picture behind the proof

The chord from $(a, f(a))$ to $(b, f(b))$ has slope $\frac{f(b) - f(a)}{b - a}$ . Subtract the chord from the graph: what is left, $g(x) = f(x) - (\text{chord at } x)$ , starts and ends at 0. A continuous function on $[a, b]$ has a maximum and a minimum (1.3); if $g$ is not constant, one of them is inside, where $g' = 0$ (Fermat), meaning $f'$ equals the chord’s slope. Picture driving 100 km in one hour: at some instant the speedometer read exactly 100 km/h.

ProofAbbott 5.3.1 to 5.3.2

The idea: Subtract the chord: g(x) = f(x) − f(a) − m(x − a) with m the chord slope. Then g(a) = g(b) = 0, the extreme value theorem gives an extremum, and Fermat makes g' vanish there.

Let

m = \frac{f(b) - f(a)}{b - a}

and

g(x) = f(x) - f(a) - m(x - a)

Name the chord slope.

g(a) = 0

and

g(b) = f(b) - f(a) - m(b - a) = 0

Substitute

x = a

and

x = b

g

attains a maximum and a minimum on

[a, b]

g

is continuous on

[a, b]

; extreme value theorem (1.3).

g

is constant, any

c \in (a, b)

has

g'(c) = 0

If both occur at the endpoints, max = min = 0 and

g

is constant 0, so

g' = 0

everywhere inside.

Otherwise an extremum occurs at some interior

c

, and

g'(c) = 0

Otherwise the max or the min is non-zero, so it is not at an endpoint (where

g = 0

); interior extremum theorem.

f'(c) = m

\blacksquare

g'(x) = f'(x) - m

Proof workshopRebuild the mean value theorem proof

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Quick check

f(x) = x^3

has

f'(0) = 0

. Does

f

have a maximum or minimum at 0?

Worked example

Find the number

c

promised by the mean value theorem for

f(x) = x^3

[0, 2]

Solution.

Check the hypotheses. $x^3$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$ . The theorem applies only when both hold.
Chord slope. $\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2} = 4$ . This is the average rate of change over the interval.
Solve $f'(c) = 4$ . $3c^2 = 4$ , so $c = \pm\frac2{\sqrt3}$ . The derivative is $3x^2$ . We set it equal to the chord slope.
Keep the root in the open interval. $\frac2{\sqrt3} \approx 1.155 \in (0, 2)$ ; the root $-1.155$ is outside. The theorem promises a $c$ inside $(a, b)$ only.

Answer. $c = \frac2{\sqrt3} \approx 1.155$ . $\blacksquare$ At that instant the tangent is parallel to the chord.

Your answerFor

f(x) = x^3

[0, 1]

, the mean value theorem gives

c \in (0, 1)

with

f'(c) =

the chord slope. Find

c

Worked example

Use the mean value theorem to prove

\ln(1 + x) \le x

for all

x > 0

Solution.

Pick the function and interval. $f(t) = \ln(1 + t)$ on $[0, x]$ . The inequality compares $f(x) - f(0)$ with $x - 0$ , which is a chord, so this is a job for the mean value theorem.
Apply the theorem. $f$ is continuous on $[0, x]$ and differentiable on $(0, x)$ , so $\ln(1 + x) - \ln 1 = f'(c)\,(x - 0)$ for some $c \in (0, x)$ . Here $f'(t) = \frac1{1 + t}$ .
Bound the derivative. $f'(c) = \frac1{1 + c} < 1$ since $c > 0$ . We cannot find $c$ , but we do not need it: any $c$ in the interval gives a derivative below 1.
Conclude. $\ln(1 + x) = f'(c)\,x < x$ . Multiply the inequality $f'(c) < 1$ by $x > 0$ .

Answer. $\ln(1 + x) < x$ for $x > 0$ (so certainly $\le$ ). $\blacksquare$ This is the standard way to use the theorem: bound $f'$ on the whole interval.

Corollary B Abbott Cor 5.3.3 and 5.3.4

Zero derivative means constant

If $f'(x) = 0$ for all $x$ in an interval, then $f$ is constant on it. Hence two functions with the same derivative on an interval differ by a constant.

In one lineIf $f'(x) = 0$ for every x in an interval, f is constant there.

Why it is trueBetween any two points, the mean value theorem finds a point where f' equals the chord slope; that slope must be 0.

ProofAbbott 5.3.4

The idea: Apply the mean value theorem on [x, y]: (f(y) − f(x))/(y − x) = f'(c) = 0.

For

x < y

in the interval,

f(y) - f(x) = f'(c)(y - x)

for some

c \in (x, y)

Any two points of the interval; MVT on

[x, y]

f(y) = f(x)

. Since

x, y

were arbitrary,

f

is constant. For two functions, apply this to their difference.

\blacksquare

f'(c) = 0

Same idea elsewhere· This is the '+ C' of chapter 2.3

Chapter 2.3 explained " $+\,C$ " by saying the null space of differentiation is the constant functions. That claim is this corollary: $f' = 0$ forces $f$ constant. It needs the mean value theorem, so it needs completeness. On the rationals, a function can have derivative 0 everywhere without being constant.

Quick checkWhy does the mean value theorem need

f

continuous on the CLOSED interval

[a, b]

Taylor’s theorem

The tangent line is the best degree-1 approximation. The best degree- $n$ approximation matches $n$ derivatives, and Taylor’s theorem says exactly how good it is.

Definition B

Taylor polynomial

If $f$ has $n$ derivatives at $a$ , its Taylor polynomial of degree $n$ at $a$ is

T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x - a)^n.

It is the only polynomial of degree at most $n$ whose value and first $n$ derivatives at $a$ agree with $f$ ‘s. (Here $k! = 1 \cdot 2 \cdots k$ , the product of the first $k$ whole numbers, and $0! = 1$ . The $k!$ in the denominator is exactly what makes the $k$ -th derivative of $\frac{(x - a)^k}{k!}$ equal to 1.)

In one line $T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x - a)^k$ : the polynomial of degree n with the same value and first n derivatives as f at a.

Why it existsIt is the best polynomial approximation near a, and it is how calculators compute sin, exp and log.

Worked example

Find the Taylor polynomial

T_3

f(x) = \ln(1 + x)

a = 0

Solution.

List the derivatives. $f'(x) = \frac1{1 + x}$ , $f''(x) = -\frac1{(1 + x)^2}$ , $f'''(x) = \frac2{(1 + x)^3}$ . $T_3$ needs $f$ and its first three derivatives.
Evaluate at $a = 0$ . $f(0) = 0$ , $f'(0) = 1$ , $f''(0) = -1$ , $f'''(0) = 2$ . The formula uses the derivatives at the centre $a$ .
Divide by factorials. The coefficients are $\frac{f^{(k)}(0)}{k!}$ : $0$ , $1$ , $-\frac12$ , $\frac26 = \frac13$ . Definition of the Taylor polynomial.
Assemble. $T_3(x) = x - \frac{x^2}2 + \frac{x^3}3$ . Sum of coefficient times $(x - 0)^k$ .

Answer. $T_3(x) = x - \frac{x^2}{2} + \frac{x^3}{3}$ . $\blacksquare$ Check: $\ln 1.1 = 0.09531$ and $T_3(0.1) = 0.1 - 0.005 + 0.000333 = 0.09533$ .

Your answerThe Taylor polynomial of

\cos x

at 0 begins

1 - \frac{x^2}{2} + \dots

. What is the coefficient of

x^4

Theorem B Abbott Thm 6.6.3, Spivak ch. 20

Taylor's theorem with Lagrange remainder

If $f$ has $n + 1$ derivatives on an interval containing $a$ and $x$ , then for some $c$ between $a$ and $x$ :

f(x) = T_n(x) + \frac{f^{(n+1)}(c)}{(n + 1)!}(x - a)^{n+1}.

In one lineIf f has n + 1 derivatives, f(x) = T_n(x) + f^{(n+1)}(c)(x − a)^{n+1}/(n+1)! for some c between a and x.

Why it is trueApply Rolle's theorem n + 1 times to the error, which vanishes to order n + 1 at a.

ProofSpivak ch. 20

The idea: Fix x. Choose the constant M so that F(t) = f(t) − T_n(t) − M(t − a)^{n+1} vanishes at t = x. F and its first n derivatives vanish at a. Rolle's theorem, applied n + 1 times, finds a point c where F^{(n+1)}(c) = 0, and that equation says M = f^{(n+1)}(c)/(n+1)!.

Let

F(t) = f(t) - T_n(t) - M(t - a)^{n+1}

, with

M

chosen so that

F(x) = 0

Fix

x \ne a

and choose

M

to make

F(x) = 0

(possible since

(x - a)^{n+1} \ne 0

F(a) = F'(a) = \dots = F^{(n)}(a) = 0

T_n

was built to match

f

's first

n

derivatives at

a

, and every derivative of

(t - a)^{n+1}

below order

n + 1

vanishes at

a

There is

c_1

between

a

and

x

with

F'(c_1) = 0

Rolle on

[a, x]

F(a) = F(x) = 0

Continuing, there are

c_2, \dots, c_{n+1}

, each between

a

and the previous one, with

F^{(k)}(c_k) = 0

Rolle again on

[a, c_1]

for

F'

, which vanishes at both ends; repeat.

0 = F^{(n+1)}(c_{n+1}) = f^{(n+1)}(c_{n+1}) - M(n + 1)!

T_n

has degree

n

, so

T_n^{(n+1)} = 0

, and the

(n+1)

-th derivative of

(t - a)^{n+1}

(n + 1)!

M = \frac{f^{(n+1)}(c)}{(n+1)!}

, and

F(x) = 0

is exactly the formula.

\blacksquare

Solve for

M

, set

c = c_{n+1}

, and use

F(x) = 0

Worked example

Which degree

n

makes the Taylor polynomial of

e^x

at 0 give

e

to within

10^{-6}

Solution.

Write the remainder. For $f = e^x$ , $a = 0$ , $x = 1$ : $e = T_n(1) + \frac{e^c}{(n + 1)!}$ for some $c \in (0, 1)$ . Every derivative of $e^x$ is $e^x$ .
Bound it. $e^c < e < 3$ , so the error is less than $\frac{3}{(n + 1)!}$ . Over-estimate freely, as in 1.2.
Solve. $\frac{3}{(n + 1)!} < 10^{-6}$ needs $(n + 1)! > 3 \times 10^6$ . Since $10! = 3628800 > 3 \times 10^6$ and $9! = 362880 < 3 \times 10^6$ , the smallest $n$ with $n + 1 \ge 10$ is $n = 9$ . Factorials grow fast; check a few values. This bound is sufficient, so a smaller $n$ might also work, but this one is guaranteed.

Answer. $T_9(1) = 1 + 1 + \frac12 + \dots + \frac1{9!} \approx 2.7182815$ , within $10^{-6}$ of $e = 2.7182818\dots$ (the true error is about $3 \times 10^{-7}$ ).

Your answerUsing the same bound

\frac{3}{(n+1)!}

, what is the smallest degree

n

that guarantees an error below

10^{-8}

Worked example

Show

|\sin x - x| \le \frac{|x|^3}{6}

for all

x

, and use it to bound the error of

\sin 0.1 \approx 0.1

Solution.

Taylor at 0 with $n = 2$ . $T_2(x) = 0 + x + 0 \cdot x^2 = x$ , since $\sin 0 = 0$ , $\cos 0 = 1$ , $-\sin 0 = 0$ . The degree-2 polynomial happens to be $x$ .
Remainder. $\sin x - x = \frac{-\cos c}{3!}x^3$ . The third derivative of $\sin$ is $-\cos$ .
Bound. $|\cos c| \le 1$ , so $|\sin x - x| \le \frac{|x|^3}{6}$ . For $x = 0.1$ : error $\le \frac{0.001}{6} \approx 0.000167$ . Actual error: $0.1 - \sin 0.1 \approx 0.000167$ , very close to the bound.

Answer. The small-angle approximation $\sin x \approx x$ used in physics (pendulums, optics) has error at most $\frac{|x|^3}6$ .

Your answerUse

|\sin x - x| \le \frac{|x|^3}{6}

x = 0.5

. What bound on the error of

\sin 0.5 \approx 0.5

do you get? (Give a fraction.)

Worked example

Estimate

\sqrt{26}

with the tangent line of

\sqrt x

at 25, and bound the error.

Solution.

Tangent line. $f(x) = \sqrt x$ , $f(25) = 5$ , $f'(x) = \frac1{2\sqrt x}$ , so $f'(25) = \frac1{10}$ . Then $\sqrt{26} \approx 5 + \frac1{10}(26 - 25) = 5.1$ . This is $T_1$ : the best straight-line estimate near 25.
Remainder formula, $n = 1$ . $\sqrt{26} - 5.1 = \frac{f''(c)}{2!}(26 - 25)^2$ for some $c \in (25, 26)$ . Taylor’s theorem with Lagrange remainder. We cannot find $c$ , so we bound the formula for every possible $c$ .
Bound $f''$ . $f''(x) = -\frac14 x^{-3/2}$ . For $c \ge 25$ , $c^{3/2} \ge 125$ , so $|f''(c)| \le \frac1{4 \cdot 125} = \frac1{500}$ . $x^{-3/2}$ gets smaller as $x$ grows, so its largest value on $(25, 26)$ is at the left end.
Error bound. $|\sqrt{26} - 5.1| \le \frac12 \cdot \frac1{500} \cdot 1 = 0.001$ . Substitute into the remainder.

Answer. $\sqrt{26} \approx 5.1$ with error at most $0.001$ . $\blacksquare$ The true value is $5.09902$ , error $0.00098$ .

Quick checkThe Taylor remainder says the error of

T_n

\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

. What is

c

Where it is used

Used inside maths· Inside maths: optimisation and Unit 5

Fermat’s theorem is why you set $f' = 0$ to find maxima; the second-derivative test is Taylor’s theorem with $n = 1$ . In several variables (5.6) the second-derivative test becomes “look at the eigenvalues of the Hessian matrix”, combining this chapter with 2.8.

Used in the real world· Physics and computing: approximations you rely on daily

The pendulum equation $\theta'' = -\frac gL \sin\theta$ is replaced by $\theta'' = -\frac gL \theta$ , using $\sin\theta \approx \theta$ with the Taylor error bound above; for a 5° swing ( $0.087$ rad) the relative error is under 0.13%. Floating-point libraries compute $\sin$ , $\exp$ and $\log$ with polynomial approximations derived from Taylor’s theorem and its refinements, choosing the degree from exactly the kind of remainder bound used in the $e$ example.

Traps

Watch out· Continuous does not mean differentiable

$|x|$ is continuous at 0 and not differentiable there. Weierstrass’s function is continuous everywhere and differentiable nowhere.

Watch out· f' = 0 at a point does not mean a maximum

$x^3$ has derivative 0 at 0 and no maximum or minimum there. Fermat’s theorem is “max implies $f' = 0$ ”, not the converse (0.1).

Watch out· The mean value theorem's c is not computable from the theorem

It guarantees a point exists. To use it, bound $f'$ on the whole interval rather than trying to find $c$ .

Problems

check

From the definition, find $f'(a)$ for $f(x) = 3x + 7$ .

Solution

$\frac{3(a + h) + 7 - 3a - 7}{h} = 3$ , so $f'(a) = 3$ .

check

Verify the mean value theorem for $f(x) = x^2$ on $[1, 3]$ : find $c$ .

Solution

Chord slope $\frac{9 - 1}{2} = 4$ ; $f'(c) = 2c = 4$ gives $c = 2 \in (1, 3)$ .

check

Write the Taylor polynomial $T_3$ of $e^x$ at 0.

Solution

$1 + x + \frac{x^2}{2} + \frac{x^3}{6}$ .

use

From the definition, prove $f(x) = \frac1x$ has $f'(a) = -\frac1{a^2}$ for $a \ne 0$ .

Solution

$\frac{\frac1{a+h} - \frac1a}{h} = \frac{a - (a + h)}{h\,a(a + h)} = \frac{-1}{a(a + h)} \to -\frac1{a^2}$ , using continuity of $\frac1{a(a + h)}$ in $h$ at 0 (1.3) for $a \ne 0$ .

use

Use the mean value theorem to prove $|\sin x - \sin y| \le |x - y|$ for all $x, y$ .

Solution

By the MVT, $\sin x - \sin y = \cos c\,(x - y)$ for some $c$ ; since $|\cos c| \le 1$ , $|\sin x - \sin y| \le |x - y|$ . (So $\sin$ is uniformly continuous, with $\delta = \varepsilon$ .)

use

Estimate $\sqrt{4.1}$ with the tangent line of $\sqrt x$ at 4, and bound the error with Taylor’s theorem.

Solution

$f(4) = 2$ , $f'(4) = \frac14$ , so $\sqrt{4.1} \approx 2 + \frac{0.1}4 = 2.025$ . Remainder: $\frac{f''(c)}{2}(0.1)^2$ with $f''(c) = -\frac14 c^{-3/2}$ , $c \in (4, 4.1)$ ; $|f''(c)| \le \frac14 \cdot \frac18 = \frac1{32}$ . Error $\le \frac1{64}(0.01) \approx 0.000156$ . (True: $\sqrt{4.1} = 2.024846$ , error $0.000154$ .)

proveAbbott Thm 5.2.4 (quotient rule)

Prove the quotient rule: if $f$ and $g$ are differentiable at $a$ and $g(a) \ne 0$ , then $\left(\frac fg\right)'(a) = \frac{f'(a)g(a) - f(a)g'(a)}{g(a)^2}$ . You may use the product rule (worked above), the chain rule, and the result $\left(\frac1x\right)' = -\frac1{x^2}$ from the previous problem.

Solution

Let $r(u) = \frac1u$ , so $\frac1g = r \circ g$ . Since $g$ is differentiable at $a$ , it is continuous there, so $g \ne 0$ near $a$ and $r \circ g$ is defined near $a$ . The chain rule with $r'(u) = -\frac1{u^2}$ gives $\left(\frac1g\right)'(a) = -\frac{g'(a)}{g(a)^2}$ . Now write $\frac fg = f \cdot \frac1g$ and use the product rule: $\left(\frac fg\right)'(a) = f'(a)\frac1{g(a)} + f(a)\left(-\frac{g'(a)}{g(a)^2}\right) = \frac{f'(a)g(a) - f(a)g'(a)}{g(a)^2}$ .

prove

Prove: if $f'(x) > 0$ for all $x$ in an interval, then $f$ is strictly increasing there.

Solution

For $x < y$ , the MVT gives $f(y) - f(x) = f'(c)(y - x) > 0$ , since both factors are positive.

prove

Prove $e^x \ge 1 + x$ for all real $x$ , using Taylor’s theorem.

Solution

$e^x = 1 + x + \frac{e^c}{2}x^2$ for some $c$ between 0 and $x$ . The last term is $\ge 0$ , so $e^x \ge 1 + x$ .

stretchAbbott Exercise 5.2.10 style

Let $f(x) = x^2 \sin(1/x)$ for $x \ne 0$ and $f(0) = 0$ . Show $f$ is differentiable at 0 with $f'(0) = 0$ , but $f'$ is not continuous at 0.

Solution

$\left|\frac{f(h) - 0}{h}\right| = |h \sin(1/h)| \le |h| \to 0$ , so $f'(0) = 0$ . For $x \ne 0$ : $f'(x) = 2x\sin(1/x) - \cos(1/x)$ . Along $x_n = \frac1{2\pi n} \to 0$ , $f'(x_n) = -1$ , not tending to $f'(0) = 0$ . So $f'$ exists everywhere but is discontinuous at 0.

stretch

L’Hospital’s rule (one case). Suppose $f(a) = g(a) = 0$ , $f, g$ differentiable near $a$ , and $g'(a) \ne 0$ . Prove $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}$ .

Solution

$\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{\frac{f(x) - f(a)}{x - a}}{\frac{g(x) - g(a)}{x - a}}$ . The numerator tends to $f'(a)$ , the denominator to $g'(a) \ne 0$ , so the quotient tends to $\frac{f'(a)}{g'(a)}$ (quotient law). (The general version with limits of $\frac{f'}{g'}$ uses the Cauchy mean value theorem.)

Summary

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ . Equivalently, $f(a + h) = f(a) + f'(a)h + E(h)$ with $\frac{E(h)}h \to 0$ : near $a$ , $f$ is a line plus a tiny error.
Differentiable implies continuous; not conversely ( $|x|$ ).
The chain rule: linear approximations compose by multiplying slopes. The proof patches the difference quotient at 0.
Fermat: interior max or min implies $f' = 0$ . Mean value theorem: some $c$ has $f'(c)$ = chord slope. Consequence: $f' = 0$ on an interval implies constant (the " $+\,C$ ").
Taylor’s theorem: $f(x) = T_n(x) + \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}$ ; bound the remainder over all $c$ to control the error.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

1.4 The derivative as best linear approximation

The picture: zoom until it is straight

Definitions

Derivative

Differentiable

Differentiable implies continuous

The rules, from the definition

Chain rule

The mean value theorem

Interior extremum theorem

Mean value theorem

Zero derivative means constant

Taylor’s theorem

Taylor polynomial

Taylor's theorem with Lagrange remainder

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere