
5.1: Derivatives of Functions of One Real Variable


In this chapter, $$"E"$$ will always denote any one of $$E^{1}, E^{*}, C$$ (the complex field$$), E^{n},^{*}$$ or another normed space. We shall consider functions $$f : E^{1} \rightarrow E$$ of one real variable with values in $$E$$ . Functions $$f : E^{1} \rightarrow E^{*}$$ (admitting finite and infinite values) are said to be extended real. Thus $$f : E^{1} \rightarrow E$$ may be real, extended real, complex, or vector valued.

Operations in $$E^{*}$$ were defined in Chapter 4, §4. Recall, in particular, our conventions $$\left(2^{*}\right)$$ there. Due to them, addition, subtraction, and multiplication are always defined in $$E^{*}$$ (with sums and products possibly "unorthodox").

To simplify formulations, we shall also adopt the convention that

$f(x)=0 \text { unless defined otherwise.}$

$$\left("0" \text { stands also for the zero-vector in } E \text { if } E \text { is a vector space.) Thus each }\right.$$ function $$f$$ is defined on all of $$E^{1}$$ . For convenience, we call $$f(x)$$ "finite" if $$f(x) \neq \pm \infty(\text { also if it is a vector })$$.

Definition

For each function $$f : E^{1} \rightarrow E,$$ we define its derived function $$f^{\prime} : E^{1} \rightarrow E$$
by setting, for every point $$p \in E^{1}$$,

$f^{\prime}(p)=\left\{\begin{array}{l}{\lim _{x \rightarrow p} \frac{f(x)-f(p)}{x-p} \text { if this limit exists (finite or not); }} \\ {0, \text { otherwise. }}\end{array}\right.$

Thus $$f^{\prime}(p)$$ is always defined.

If the limit in $$(1)$$ exists, we call it the derivative of $$f$$ at $$p$$.

If, in addition, this limit is finite, we say that $$f$$ is differentiable at $$p$$.

If this holds for each $$p$$ in a set $$B \subseteq E^{1},$$ we say that $$f$$ has a derivative (respectively, is differentiable) on $$B,$$ and we call the function $$f^{\prime}$$ thederivative of $$f$$ on $$B$$.

If the limit in $$(1)$$ is one sided (with $$x \rightarrow p^{-}$$ or $$x \rightarrow p^{+} ),$$ we call it a one-sided (left or right) derivative at $$p,$$ denoted $$f_{-}^{\prime}$$ or $$f_{+}^{\prime}$$.

Definition

Given a function $$f : E^{1} \rightarrow E,$$ we define its $$n$$ th derived function (or derived function of order $$n ),$$ denoted $$f^{(n)} : E^{1} \rightarrow E,$$ by induction:

$f^{(0)}=f, f^{(n+1)}=\left[f^{(n)}\right]^{\prime}, \quad n=0,1,2, \ldots$

Thus $$f^{(n+1)}$$ is the derived function of $$f^{(n)} .$$ By our conventions, $$f^{(n)}$$ is defined on all of $$E^{1}$$ for each $$n$$ and each function $$f : E^{1} \rightarrow E .$$ We have $$f^{(1)}=f^{\prime},$$ and we write $$f^{\prime \prime}$$ for $$f^{(2)}, f^{\prime \prime \prime}$$ for $$f^{(3)},$$ etc. We say that $$f$$ has $$n$$ derivatives at a point $$p$$ iff the limits

$\lim _{x \rightarrow q} \frac{f^{(k)}(x)-f^{(k)}(q)}{x-q}$

exist for all $$q$$ in a neighborhood $$G_{p}$$ of $$p$$ and for $$k=0,1, \ldots, n-2,$$ and also

$\lim _{x \rightarrow p} \frac{f^{(n-1)}(x)-f^{(n-1)}(p)}{x-p}$

exists. If all these limits are finite, we say that $$f$$ is $$n$$ times differentiable on $$I ;$$ similarly for one-sided derivatives.

It is an important fact that differentiability implies continuity.

Theorem $$\PageIndex{1}$$

If a function $$f : E^{1} \rightarrow E$$ is differentiable at a point $$p \in E^{1},$$ it is continuous at $$p,$$ and $$f(p)$$ is finite (even if $$E=E^{*} )$$.

Proof

Setting $$\Delta x=x-p$$ and $$\Delta f=f(x)-f(p),$$ we have the identity

$|f(x)-f(p)|=\left|\frac{\Delta f}{\Delta x} \cdot(x-p)\right| \quad \text { for } x \neq p.$

By assumption,

$f^{\prime}(p)=\lim _{x \rightarrow p} \frac{\Delta f}{\Delta x}$

exists and is finite. Thus as $$x \rightarrow p,$$ the right side of $$(2)$$ (hence the left side as well) tends to $$0,$$ so

$\lim _{x \rightarrow p}|f(x)-f(p)|=0, \text { or } \lim _{x \rightarrow p} f(x)=f(p)$

proving continuity at $$p$$.

Also, $$f(p) \neq \pm \infty,$$ for otherwise $$|f(x)-f(p)|=+\infty$$ for all $$x,$$ and so $$|f(x)-f(p)|$$ cannot tend to $$0 . \quad \square$$

Note 1. Similarly, the existence of a finite left (right) derivative at $$p$$ implies left (right) continuity at $$p$$ . The proof is the same.

Note 2. The existence of an infinite derivative does not imply continuity, nor does it exclude it. For example, consider the two cases

(i) $$f(x)=\frac{1}{x},$$ with $$f(0)=0,$$ and

(ii) $$f(x)=\sqrt[3]{x}$$.

Give your comments for $$p=0$$.

Caution: A function may be continuous on $$E^{1}$$ without being differentiable anywhere (thus the converse to Theorem 1 fails). The first such function was indicated by Weierstrass. We give an example due to Olmsted (Advanced Calculus).

Example $$\PageIndex{1}$$

(a) We first define a sequence of functions $$f_{n} : E^{1} \rightarrow E^{1}(n=1,2, \ldots)$$ as follows. For each $$k=0, \pm 1, \pm 2, \ldots,$$ let

$f_{n}(x)=0 \text { if } x=k \cdot 4^{-n}, \text { and } f_{n}(x)=\frac{1}{2} \cdot 4^{-n} \text { if } x=\left(k+\frac{1}{2}\right) \cdot 4^{-n}.$

Between $$k \cdot 4^{-n}$$ and $$\left(k \pm \frac{1}{2}\right) \cdot 4^{-n}, f_{n}$$ is linear (see Figure $$21 ),$$ so it is continuous on $$E^{1} .$$ The series $$\sum f_{n}$$ converges uniformly on $$E^{1} .$$ (Verify!)

Let

$f=\sum_{n=1}^{\infty} f_{n}.$

Then $$f$$ is continuous on $$E^{1}(\text { why? yet it is nowhere differentiable.}$$

To prove this fact, fix any $$p \in E^{1} .$$ For each $$n,$$ let

$x_{n}=p+d_{n}, \text { where } d_{n}=\pm 4^{-n-1},$

choosing the sign of $$d_{n}$$ so that $$p$$ and $$x_{n}$$ are in the same half of a "sawtooth" in the graph of $$f_{n}$$ (Figure 21$$)$$ . Then

$f_{n}\left(x_{n}\right)-f_{n}(p)=\pm d_{n}=\pm\left(x_{n}-p\right) . \quad(\text { Why } ?)$

Also,

$f_{m}\left(x_{n}\right)-f_{m}(p)=\pm d_{n} \text { if } m \leq n$

but vanishes for $$m>n .$$ (Why?)

Thus, when computing $$f\left(x_{n}\right)-f(p),$$ we may replace

$f=\sum_{m=1}^{\infty} f_{m} \text { by } f=\sum_{m=1}^{n} f_{m}.$

Since

$\frac{f_{m}\left(x_{n}\right)-f_{m}(p)}{x_{n}-p}=\pm 1 \text { for } m \leq n.$

the fraction

$\frac{f\left(x_{n}\right)-f(p)}{x_{n}-p}$

is an integer, odd if $$n$$ is odd and even if $$n$$ is even. Thus this fraction cannot tend to a finite limit as $$n \rightarrow \infty,$$ i.e., as $$d_{n}=4^{-n-1} \rightarrow 0$$ and $$x_{n}=p+d_{n} \rightarrow p .$$ A fortiori, this applies to

$\lim _{x \rightarrow p} \frac{f(x)-f(p)}{x-p}.$

Thus $$f$$ is not differentiable at any $$p$$.

The expressions $$f(x)-f(p)$$ and $$x-p,$$ briefly denoted $$\Delta f$$ and $$\Delta x,$$ and $$\Delta x,$$ are called the increments of $$f$$ and $$x$$ (at $$p ),$$ respectively. 2 We now show that for differentiable functions, $$\Delta f$$ and $$\Delta x$$ are "nearly proportional' when $$x$$ approaches $$p ;$$ that is,

$\frac{\Delta f}{\Delta x}=c+\delta(x)$

with $$c$$ constant and $$\lim _{x \rightarrow p} \delta(x)=0$$.

Theorem $$\PageIndex{2}$$

A function $$f : E^{1} \rightarrow E$$ is differentiable at $$p,$$ and $$f^{\prime}(p)=c,$$ iff there is a finite $$c \in E$$ and a function $$\delta : E^{1} \rightarrow E$$ such that $$\lim _{x \rightarrow p} \delta(x)=\delta(p)=0,$$ and such that

$\Delta f=[c+\delta(x)] \Delta x \quad \text { for all } x \in E^{1}.$

Proof

If $$f$$ is differentiable at $$p,$$ put $$c=f^{\prime}(p) .$$ Define $$\delta(p)=0$$ and

$\delta(x)=\frac{\Delta f}{\Delta x}-f^{\prime}(p) \text { for } x \neq p.$

Then $$\lim _{x \rightarrow p} \delta(x)=f^{\prime}(p)-f^{\prime}(p)=0=\delta(p) .$$ Also, $$(3)$$ follows.

Conversely, if $$(3)$$ holds, then

$\frac{\Delta f}{\Delta x}=c+\delta(x) \rightarrow c \text { as } x \rightarrow p(\text { since } \delta(x) \rightarrow 0).$

Thus by definition,

$c=\lim _{x \rightarrow p} \frac{\Delta f}{\Delta x}=f^{\prime}(p) \text { and } f^{\prime}(p)=c \text { is finite. } \square$

Theorem $$\PageIndex{3}$$

(chain rule). Let the functions $$g : E^{1} \rightarrow E^{1}(\text { real })$$ and $$f : E^{1} \rightarrow E$$ (real or not) be differentiable at $$p$$ and $$q,$$ respectively, where $$q=g(p) .$$ Then the composite function $$h=f \circ g$$ is differentiable at $$p,$$ and

$h^{\prime}(p)=f^{\prime}(q) g^{\prime}(p).$

Proof

Setting

$\Delta h=h(x)-h(p)=f(g(x))-f(g(p))=f(g(x))-f(q).$

we must show that

$\lim _{x \rightarrow p} \frac{\Delta h}{\Delta x}=f^{\prime}(q) g^{\prime}(p) \neq \pm \infty.$

Now as $$f$$ is differentiable at $$q,$$ Theorem 2 yields a function $$\delta : E^{1} \rightarrow E$$ such that $$\lim _{x \rightarrow q} \delta(x)=\delta(q)=0$$ and such that

$\left(\forall y \in E^{1}\right) \quad f(y)-f(q)=\left[f^{\prime}(q)+\delta(y)\right] \Delta y, \Delta y=y-q.$

Taking $$y=g(x),$$ we get

$\left(\forall x \in E^{1}\right) \quad f(g(x))-f(q)=\left[f^{\prime}(q)+\delta(g(x))\right][g(x)-g(p)],$

where

$g(x)-g(p)=y-q=\Delta y \text { and } f(g(x))-f(q)=\Delta h,$

as noted above. Hence

$\frac{\Delta h}{\Delta x}=\left[f^{\prime}(q)+\delta(g(x))\right] \cdot \frac{g(x)-g(p)}{x-p} \quad \text { for all } x \neq p.$

Let $$x \rightarrow p .$$ Then we obtain $$h^{\prime}(p)=f^{\prime}(q) g^{\prime}(p),$$ for, by the continuity of $$\delta \circ g$$ at $$p$$ (Chapter 4, §2, Theorem 3),

$\lim _{x \rightarrow p} \delta(g(x))=\delta(g(p))=\delta(q)=0 . \square$

The proofs of the next two theorems are left to the reader.

Theorem $$\PageIndex{4}$$

If $$f, g,$$ and $$h$$ are real or complex and are differentiable at $$p,$$ so are

$f \pm g, h f, \text { and } \frac{f}{h}$

(the latter if $$h(p) \neq 0 ),$$ and at the point $$p$$ we have

(i) $$(f \pm g)^{\prime}=f^{\prime} \pm g^{\prime}$$;

(ii) $$(h f)^{\prime}=h f^{\prime}+h^{\prime} f ;$$ and

(iii) $$\left(\frac{f}{h}\right)^{\prime}=\frac{h f^{\prime}-h^{\prime} f}{h^{2}}$$.

All this holds also if $$f$$ and $$g$$ are vector valued and $$h$$ is scalar valued. It also applies to infinite (even one-sided) derivatives, except when the limits involved become indeterminate (Chapter 4, §4).

Note 3. By induction, if $$f, g,$$ and $$h$$ are $$n$$ times differentiable at a point $$p,$$ so are $$f \pm g$$ and $$h f,$$ and, denoting by $$\left(\begin{array}{l}{n} \\ {k}\end{array}\right)$$ the binomial coefficients, we have

(i*) $$(f \pm g)^{(n)}=f^{(n)} \pm g^{(n)} ;$$ and

(ii*) $$(h f)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l}{n} \\ {k}\end{array}\right) h^{(k)} f^{(n-k)}$$.

Formula (ii $$)$$ is known as the Leibniz formula; its proof is analogous to that of the binomial theorem. It is symbolically written as $$(h f)^{(n)}=(h+f)^{n},$$ with the last term interpreted accordingly.

Theorem $$\PageIndex{5}$$

(componentwise differentiation). A function $$f : E^{1} \rightarrow E^{n}\left(^{*} C^{n}\right)$$ is differentiable at $$p$$ iff each of its $$n$$ components $$\left(f_{1}, \ldots, f_{n}\right)$$ is, and then

$f^{\prime}(p)=\left(f_{1}^{\prime}(p), \ldots, f_{n}^{\prime}(p)\right)=\sum_{k=1}^{n} f_{k}^{\prime}(p) \overline{e}_{k},$

with $$\overline{e}_{k}$$ as in Theorem 2 of Chapter 3, §§1-3.

In particular, a complex function $$f : E^{1} \rightarrow C$$ is differentiable iff its real and imaginary parts are, and $$f^{\prime}=f_{\mathrm{re}}^{\prime}+i \cdot f_{\text { im }}^{\prime}$$ Chapter 4, §3, Note 5).

Example $$\PageIndex{2}$$

(b) Consider the complex exponential

$f(x)=\cos x+i \cdot \sin x=e^{x i}(\text { Chapter } 4, §3).$

We assume the derivatives of $$\cos x$$ and $$\sin x$$ to be known (see Problem 8$$) .$$ By Theorem $$5,$$ we have

$f^{\prime}(x)=-\sin x+i \cdot \cos x=\cos \left(x+\frac{1}{2} \pi\right)+i \cdot \sin \left(x+\frac{1}{2} \pi\right)=e^{\left(x+\frac{1}{2} \pi\right) i}.$

Hence by induction,

$f^{(n)}(x)=e^{\left(x+\frac{1}{2} n \pi\right) i}, n=1,2, \ldots .(\text { Verify! })$

(c) Define $$f : E^{1} \rightarrow E^{3}$$ by

$f(x)=(1, \cos x, \sin x), \quad x \in E^{1}.$

Here Theorem 5 yields

$f^{\prime}(p)=(0,-\sin p, \cos p), \quad p \in E^{1}.$

For a fixed $$p=p_{0},$$ we may consider the line

$\overline{x}=\overline{a}+t \vec{u},$

where

$\overline{a}=f\left(p_{0}\right) \text { and } \vec{u}=f^{\prime}\left(p_{0}\right)=\left(0,-\sin p_{0}, \cos p_{0}\right).$

This is, by definition, the tangent vector at $$p_{0}$$ to the curve $$f\left[E^{1}\right]$$ in $$E^{3}$$.

More generally, if $$f : E^{1} \rightarrow E$$ is differentiable at $$p$$ and continuous on some globe about $$p,$$ we define the tangent at $$p$$ to the curve $$f\left[G_{p}\right]$$ to be the line

$\overline{x}=f(p)+t \cdot f^{\prime}(p);$

$$f^{\prime}(p)$$ is its direction vector in $$E,$$ while $$t$$ is the variable real parameter. For real functions $$f : E^{1} \rightarrow E^{1},$$ we usually consider not $$f\left[E^{1}\right]$$ but the curve $$y=f(x)$$ in $$E^{2},$$ i.e., the set

$\left\{(x, y) | y=f(x), x \in E^{1}\right\}.$

The tangent to that curve at $$p$$ is the line through $$(p, f(p))$$ with slope $$f^{\prime}(p)$$.

In conclusion, let us note that differentiation (i.e., taking derivatives) is a local limit process at some point $$p .$$ Hence (cf. Chapter 4, §1, Note 4 ) the existence and the value of $$f^{\prime}(p)$$ is not affected by restricting $$f$$ to some globe $$G_{p}$$ about $$p$$ or by arbitrarily redefining $$f$$ outside $$G_{p} .$$ For one-sided derivatives, we may replace $$G_{p}$$ by its corresponding "half."