5.1 Differentiable functions: definitions and basic examples

Definition of differentiability

Geometrically, differentiation concerns the gradient, or slope, of a graph of a function. This notion is easiest to define when the graph is just a straight line.

Example 5.1.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be the function $f(x):=x/2+1$ for all $x\in\mathbb{R}$ . If we pick two distinct points, say $a$ and $a+h$ for $a\in\mathbb{R}$ and $h\in\mathbb{R}\setminus\{0\}$ , then

f(a+h)-f(a)=\frac{a+h}{2}+1-\frac{a}{2}-1=\frac{h}{2}

is the vertical change of the function between $a$ and $a+h$ and $h=(a+h)-a$ is the horizontal change: see Figure 5.3. The gradient of $f$ is then defined as the ratio of the vertical change and the horizontal change:

\frac{f(a+h)-f(a)}{h}=\frac{h}{2h}=\frac{1}{2}.

Note that we get the same value $1/2$ no matter which values of $a$ and $h$ we choose. This is because the graph of $f$ is a straight line, so has constant slope everywhere.

Figure 5.3: Computing the gradient of the line

f(x):=x/2+1

If the graph of our function is curved, then the gradient will no longer be a constant value. In this case, when we talk about the gradient of $f$ at a point $a$ , what we really wish to describe is the gradient of the tangent line to the function at the point $a$ (if such a tangent line exists). But what really is a ‘tangent line’? How do we define it?

For fixed $a\in\mathbb{R}$ and $h\in\mathbb{R}\setminus\{0\}$ , we can still consider the difference quotient

(5.1) (5.1)

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

This corresponds to the gradient of the secant line passing through the points $(a,f(a))$ and $(a+h,f(a+h))$ on the graph of $f$ : see Figure 5.4.

Figure 5.4: The gradient of the secant line through

(a,f(a))

and

(a+h,f(a+h))

. The tangent line to the curve at

(a,f(a))

is shown as a dashed purple line.

In general, the secant line doesn’t look very much like a tangent line; indeed, this is the case in Figure 5.4, where we see the secant line is quite far from the tangent line. However, if $h$ is very small, then we hope that the secant line gives a reasonable approximation to a tangent line at $a$ . Moreover, as $h$ gets smaller and smaller, we can hope for a better and better approximation. This idea is illustrated in Figure 5.5. In this case, as $h$ gets smaller and smaller, the difference quotient (5.1) gives a better and better approximation to the slope or gradient of the function at $a$ . This is the intuition behind the definition of the derivative.

Definition 5.2.

Let $I\subseteq\mathbb{R}$ be an open interval.

1

A function $f\colon I\to\mathbb{R}$ is said to be differentiable at $a\in I$ if

(5.2) (5.2) $f^{\prime}(a):=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

exists. The number $f^{\prime}(a)$ is called the derivative of $f$ at $a$ .
2

We say $f$ is differentiable if it is differentiable at all points $a\in I$ . In this case, we define the derivative of $f$ to be the function $f^{\prime}\colon I\to\mathbb{R}$ mapping $a$ to $f^{\prime}(a)$ for all $a\in I$ .

Figure 5.5: The definition of the derivative. We consider the secant line passing through

(a,f(a))

and

(a+h,f(a+h))

for smaller and smaller values of

h

Remark 5.3.

In the definition of the derivative, it is sometimes convenient to change variables, replacing $a+h$ with $x$ . In particular, we may equivalently write (5.2) as

f^{\prime}(a):=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.

By generalising the computation from Example 5.1, we can show that the definition of a derivative agrees with our usual notion of gradient or slope when the graph of the function is a straight line.

Exercise 5.4.

Let $m$ , $c\in\mathbb{R}$ and consider the linear function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=mx+c$ for all $x\in\mathbb{R}$ . Arguing from the definition, show that $f$ is differentiable and $f^{\prime}(a)=m$ for all $a\in\mathbb{R}$ .

The motivation for the definition of a derivative relied on our intuition about tangent lines. However, so far we still have not given a precise definition of a ‘tangent line’. Now that we have a formal definition of a derivative, we can address this.

Definition 5.5.

Let $I\subseteq\mathbb{R}$ be an open interval and $f\colon I\to\mathbb{R}$ be differentiable at $a\in I$ . We define the tangent line to the graph of $f$ at $a$ to be the line through $(a,f(a))$ with slope $f^{\prime}(a)$ .

The next example illustrates the derivative in a case where the graph of the function is curved.

Example 5.6.

Consider the function $p_{2}\colon\mathbb{R}\to\mathbb{R}$ given by $p_{2}(x):=x^{2}$ for all $x\in\mathbb{R}$ . Then $p_{2}$ is differentiable and $p_{2}^{\prime}(a)=2a$ for all $a\in\mathbb{R}$ .

You are no doubt aware of this fact from earlier courses, but our goal here is to justify the value of the derivative by arguing from the definition. Reasoning in this way gives a rigorous foundation for why the various derivative formulæ you’ve encountered hold, exploiting the tools that we have built up so far in the course.

Proof.

Fix $a\in\mathbb{R}$ and $h\in\mathbb{R}\setminus\{0\}$ . By expanding out the square, we see that

\frac{p_{2}(a+h)-p_{2}(a)}{h}=\frac{(a+h)^{2}-a^{2}}{h}=\frac{a^{2}+2ah+h^{2}-% a^{2}}{h}=\frac{2ah+h^{2}}{h}=2a+h.

Taking the limit as $h\to 0$ , we therefore see that

\lim_{h\to 0}\frac{p_{2}(a+h)-p_{2}(a)}{h}=2a

and so $p_{2}$ is differentiable at $a$ and $p_{2}^{\prime}(a)=2a$ , as required. ∎

It is equally important to identify functions which fail to be differentiable. The following functions are classic examples of functions which are continuous but are not differentiable.

Example 5.7.

The continuous function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=|x|$ (shown as $f_{1}$ in Figure 5.1) is not differentiable at $0$ .

Proof.

The one-sided limits of the difference quotient are given by

\lim_{h\to 0_{+}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0_{+}}\frac{h}{h}=1,\qquad% \lim_{h\to 0_{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0_{-}}-\frac{h}{h}=-1.

Since the one-sided limits do not agree, we know from Exercise 4.72 that $\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$ does not exist. Hence, $f$ is not differentiable at $0$ . ∎

Example 5.8.

The continuous function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=\sqrt{|x|}$ (shown as $f_{3}$ in Figure 5.1) is not differentiable at $0$ .

Proof.

For $h\in\mathbb{R}\setminus\{0\}$ , the difference quotient is given by

\frac{f(0+h)-f(0)}{h}=\frac{h^{1/2}}{h}=h^{-1/2}.

For any $\delta>0$ , the right-hand expression is unbounded on the interval $(0,\delta)$ , and so $\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$ does not exist by Exercise 4.48 ii). Hence, $f$ is not differentiable at $0$ . ∎

Exercise 5.9.

Show that the function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=\begin{cases}x^{2}&\text{if }x\leq 0\\ x&\text{if }x>0\end{cases}$ (shown as $f_{2}$ in Figure 5.1) is not differentiable at 0.

Remark 5.10.

There are different ways to denote derivatives, some of which you will likely have come across. For instance, it is common to use the notation

f^{\prime}=f^{(1)}=\frac{\mathrm{d}f}{\mathrm{d}x}=D_{x}f

interchangeably. We shall primarily use $f^{\prime}$ in this course.

Further examples of differentiable functions

We shall consider some more sophisticated examples of differentiable functions and their derivatives. Many or all of these formulæ will be familiar to you, but as before the onus is on developing a rigorous understanding of why the formulæ hold by arguing from the definition. We want to get ‘under the hood’ and really understand how and why derivatives work! Our first example generalises Example 5.6.

Example 5.11 (Monomials).

For all $k\in\mathbb{N}$ , the function $p_{k}\colon\mathbb{R}\to\mathbb{R}$ given by $p_{k}(x):=x^{k}$ for all $x\in\mathbb{R}$ is differentiable. Moreover, $p_{k}^{\prime}(a)=ka^{k-1}$ for all $a\in\mathbb{R}$ .

Note that the $k=2$ case corresponds to Example 5.6; however, here we shall use a slightly different argument to that of Example 5.6, since it will later help us to study the exponential function.

Proof.

Fix $a\in\mathbb{R}$ and $h\in\mathbb{R}$ with $0<|h|\leq 1$ . We apply the binomial theorem to write

p_{k}(a+h)-p_{k}(a)=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}h^{j}-a^{k}=ka^{k-1}h+% \sum_{j=2}^{k}\binom{k}{j}a^{k-j}h^{j},

since in the sum, the $j=0$ term is $a^{k}$ and the $j=1$ term is $ka^{k-1}h$ . This rearranges to give

(5.3) (5.3)

\frac{p_{k}(a+h)-p_{k}(a)}{h}-ka^{k-1}=h\sum_{j=2}^{k}\binom{k}{j}a^{k-j}h^{j-% 2}.

Here for $k=1$ the right-hand sum in (5.3) is interpreted as equal to $0$ . By the triangle inequality, the hypothesis $|h|\leq 1$ and a second application of the binomial theorem,

(5.4) (5.4)

\Big{|}\sum_{j=2}^{k}\binom{k}{j}a^{k-j}h^{j-2}\Big{|}\leq\sum_{j=2}^{k}\binom% {k}{j}|a|^{k-j}|h|^{j-2}\leq\sum_{j=0}^{k}\binom{k}{j}|a|^{k-j}=(1+|a|)^{k}.

Combining (5.3) and (5.4) we see that

(5.5) (5.5)

\Big{|}\frac{p_{k}(a+h)-p_{k}(a)}{h}-ka^{k-1}\Big{|}\leq|h|(1+|a|)^{k}.

Thus, we conclude from the squeeze theorem that

\lim_{h\to 0}\frac{p_{k}(a+h)-p_{k}(a)}{h}=ka^{k-1}

and so $p_{k}$ is differentiable at $a$ and $p^{\prime}(a)=ka^{k-1}$ , as required. ∎

We can use the ideas introduced in Example 5.11 to study the following very important example.

Lemma 5.12.

The function $\exp\colon\mathbb{R}\to\mathbb{R}$ introduced in Definition 4.5 is differentiable with

\exp^{\prime}(a)=\exp(a)\qquad\text{for all $a\in\mathbb{R}$.}

Lemma 5.12 tells us that the function $\exp$ is its own derivative. As a consequence, $\exp$ plays a very special role in the theory of differentiation. We shall see many important consequences of this result in Section 5.7 below.

Proof (of Lemma 5.12).

As in Example 5.11, for each $k\in\mathbb{N}_{0}$ let $p_{k}\colon\mathbb{R}\to\mathbb{R}$ be given by $p_{k}(x):=x^{k}$ for all $x\in\mathbb{R}$ . Then, given $n\in\mathbb{N}_{0}$ , let

s_{n}\colon\mathbb{R}\to\mathbb{R}\qquad\text{be given by}\qquad s_{n}(x):=% \sum_{k=0}^{n}\frac{x^{k}}{k!}=\sum_{k=0}^{n}\frac{p_{k}(x)}{k!}\qquad\text{% for $x\in\mathbb{R}$,}

so that $s_{n}(x)$ corresponds to the $n$ th partial sum of the series $\exp(x)$ for all $x\in\mathbb{R}$ . Fix $a\in\mathbb{R}$ and $h\in\mathbb{R}$ with $0<|h|\leq 1$ and consider the difference quotient

(5.6) (5.6)

\frac{s_{n}(a+h)-s_{n}(a)}{h}=\sum_{k=1}^{n}\frac{1}{k!}\cdot\frac{p_{k}(a+h)-% p_{k}(a)}{h};

notice that the sum on the right starts from $k=1$ since $p_{0}(a+h)-p_{0}(a)=0$ . On the other hand, by reindexing the sum we can write

(5.7) (5.7)

s_{n-1}(a)=\sum_{k=0}^{n-1}\frac{a^{k}}{k!}=\sum_{k=1}^{n}\frac{a^{k-1}}{(k-1)% !}=\sum_{k=1}^{n}\frac{ka^{k-1}}{k!}.

Combining (5.6) and (5.7), we see that

\frac{s_{n}(a+h)-s_{n}(a)}{h}-s_{n-1}(a)=\sum_{k=1}^{n}\frac{1}{k!}\cdot\Big{[% }\frac{p_{k}(a+h)-p_{k}(a)}{h}-ka^{k-1}\Big{]}

Notice that the right-hand side involves the difference quotients for the monomials $p_{k}$ we studied in Example 5.11. So, we can apply some of the inequalities we proved in that example.

Using the triangle inequality and the bound (5.5) from Example 5.11, we obtain

\Big{|}\frac{s_{n}(a+h)-s_{n}(a)}{h}-s_{n-1}(a)\Big{|}\leq|h|\sum_{k=1}^{n}% \frac{(1+|a|)^{k}}{k!}\leq|h|\sum_{k=0}^{\infty}\frac{(1+|a|)^{k}}{k!}=|h|\exp% (1+|a|).

Since the above inequality holds for all $n\in\mathbb{N}$ , we can take the limit as $n\to\infty$ to deduce (by Exercise 2.46) that

\Big{|}\frac{\exp(a+h)-\exp(a)}{h}-\exp(a)\Big{|}\leq|h|\exp(1+|a|).

As $h\to 0$ , the right-hand side converges to $0$ . Thus, by the squeeze theorem,

\lim_{h\to 0}\frac{\exp(a+h)-\exp(a)}{h}=\exp(a)

and so $\exp$ is differentiable at $a$ and $\exp^{\prime}(a)=\exp(a)$ , as required. ∎

Example 5.13 (Trigonometric functions).

The functions $\sin\colon\mathbb{R}\to\mathbb{R}$ and $\cos\colon\mathbb{R}\to\mathbb{R}$ are differentiable with $\sin^{\prime}(a)=\cos a$ and $\cos^{\prime}(a)=-\sin a$ for all $a\in\mathbb{R}$ .

Proof.

For $\mathrm{sinc}^{*}$ the function defined in Example 4.44, observe that

\lim_{h\to 0}\frac{\sin h-\sin 0}{h}=\lim_{h\to 0}\frac{\sin h}{h}=\lim_{h\to 0% }\mathrm{sinc}^{*}(h)=1.

This tells us that $\sin$ is differentiable at $0$ and $\sin^{\prime}(0)=1$ . A similar argument shows that $\cos$ is differentiable at $0$ and $\cos^{\prime}(0)=0$ : see Exercise 5.14 (i).

So far we have only shown $\sin$ and $\cos$ are differentiable at $0$ . However, this special case can be combined with the angle summation formulæ to prove the full result: see Exercise 5.14 (ii). ∎

Exercise 5.14.

Fill in the details of Example 5.13 by carrying out the following steps.

(i)

Use Exercise 4.46 to show that $\cos$ is differentiable at $0$ and $\cos^{\prime}(0)=0$ .
(ii)

Fix $a\in\mathbb{R}$ and use the angle summation formulæ to show that $\sin$ and $\cos$ are both differentiable at $a\in\mathbb{R}$ with $\sin^{\prime}(a)=\cos a$ and $\cos^{\prime}(a)=-\sin a$ .

Higher-order derivatives

If a function $f\colon I\to\mathbb{R}$ is differentiable, this means that the derivative $f^{\prime}\colon I\to\mathbb{R}$ exists – but the function $f^{\prime}$ may or may not be differentiable. This motivates the following definition, to distinguish the different types of behaviour that are possible.

Definition 5.15.

Let $I\subseteq\mathbb{R}$ be an open interval and $f\colon I\to\mathbb{R}$ be differentiable.

1

If $f^{\prime}$ is differentiable then we say $f$ is twice differentiable and define $f^{(2)}:=(f^{\prime})^{\prime}$ .
2

More generally, for $n\in\mathbb{N}$ we say $f\colon I\to\mathbb{R}$ is $n$ -times differentiable if the (recursively-defined) higher-order derivatives $f^{(k+1)}:=(f^{(k)})^{\prime}$ exist for all $0\leq k\leq n-1$ .
3

Finally, we say that $f$ is infinitely differentiable if it is $n$ -times differentiable for all $n\in\mathbb{N}$ .

Exercise 5.16.

Show that $\exp\colon\mathbb{R}\to\mathbb{R}$ is infinitely differentiable.

Exercise 5.17.

Consider the function $p_{2}\colon\mathbb{R}\to\mathbb{R}$ given by $p_{2}(x):=x^{2}$ for all $x\in\mathbb{R}$ . Is $p_{2}$ twice differentiable? Three times differentiable? Infinitely differentiable?

Example 5.18.

The function

(5.8) (5.8)

f\colon\mathbb{R}\to\mathbb{R},\qquad f(x):=\begin{cases}0&\text{if $x<0$,}\\ x^{2}&\text{if $x\geq 0$}\end{cases}

is differentiable but not twice differentiable. To see this, note our earlier examples show that $f$ is differentiable on $(-\infty,0)$ with derivative $f^{\prime}(a)=0$ for all $a<0$ and $f$ is is differentiable on $(0,\infty)$ with derivative $f^{\prime}(a)=2a$ for all $x>0$ . For $a=0$ , we have

\lim_{h\to 0_{+}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0_{+}}\frac{h^{2}}{h}=0,% \qquad\lim_{h\to 0_{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0_{-}}0=0.

Since the left and right limits agree, we conclude from Exercise 4.72 that $f$ is differentiable at $0$ with $f^{\prime}(0)=0$ . However, the above argument shows that

f^{\prime}(x):=\begin{cases}0&\text{if $x<0$,}\\ 2x&\text{if $x\geq 0$}\end{cases}

and it is easy to see this is not differentiable using an argument similar to that from Example 5.7. We illustrate this example in Figure 5.6.

Figure 5.6: The function

f

from (5.8) and its derivative

f^{\prime}

Remark 5.19.

Again, there are different ways to denote higher-order derivatives. These include

f^{(n)}=\frac{d^{n}f}{dx^{n}}=D^{n}_{x}f.

Furthermore, second derivatives can be written as $f^{(2)}$ or $f^{\prime\prime}$ (and similarly for higher-order derivatives, although by the time we get to fourth order the notation $f^{\prime\prime\prime\prime}$ gets out of hand).

One-sided derivatives

As with continuity, it is convenient to define one-sided derivatives to deal with functions whose domains are closed (or half-open) intervals. For instance, given $a$ , $b\in\mathbb{R}$ with $a<b$ , we say $f\colon[a,b]\to\mathbb{R}$ is differentiable on $[a,b]$ if $f$ is differentiable at all points $x\in(a,b)$ and the limits

f^{\prime}_{R}(a):=\lim_{h\to 0_{+}}\frac{f(a+h)-f(a)}{h}\qquad\text{and}% \qquad f^{\prime}_{L}(b):=\lim_{h\to 0_{-}}\frac{f(b+h)-f(b)}{h}.

exist. In this case, $f^{\prime}_{R}(a)$ is called the right derivative at $a$ and $f^{\prime}_{L}(b)$ is called left derivative at $b$ .