7.1 The Derivative

The derivative is an extremely important concept in applied mathematics, as it allows us to model real life situations with mathematical functions and their rates of change. For this reason, the derivative is the most interesting part of analysis for applied mathematicians and physicists. The purpose of this chapter is to establish what a derivative is, what its properties are, and what we can do with derivatives.

7.1.1 Differentiability

You probably have an intuition that the derivative of a function is the gradient of the tangent line at every point along the function’s domain, and that in order for a function to have a derivative defined everywhere it must have no sharp corners or cusps. This is the basic idea, but of course in analysis we must be extra careful with our definitions. How do we know if we can even define a tangent line? To make a definition of the derivative that we can work with we will use equipment that we have developed in the last chapter.

Definition 7.1.

Let $I\subseteq\mathbb{R}$ be an open interval and let $f:I\to\mathbb{R}$ be a function. Let $c\in I$ . Then $f$ is differentiable at $c$ if

\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\,\text{exists.}

(7.1)

If this limit exists, then we denote it $f^{\prime}(c)$ , the derivative of $f$ at $c$ . We say $f$ is differentiable if $f$ is differentiable at all $c\in I$ . Then $f^{\prime}:I\to\mathbb{R}$ is also a function, called the derivative of $f$ (also denoted $\frac{\mathrm{d}f}{\mathrm{d}x}$ ).

Example 7.1.

Let $f:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=x^{2}$ . Claim: $f$ is differentiable at $x=3$ with $f^{\prime}(3)=6$ .
Proof: Let $c=3,x\neq 3$ . Then

\frac{f(x)-f(3)}{x-3}=\frac{x^{2}-3^{2}}{x-3}=\frac{(x+3)(x-3)}{x-3}=x+3.

(7.2)

Hence,

f^{\prime}(3)=\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}=\lim_{x\to 3}x+3=6.

(7.3)

So $f$ is differentiable at $x=3$ with derivative 6. In fact, if $c\neq x$ ,

\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}x+c=2c.

(7.4)

So $f$ is differentiable everywhere with derivative $f^{\prime}:\mathbb{R}\to\mathbb{R}$ given by $f^{\prime}(x)=2x$ .

Example 7.2.

Let $f:(0,\infty)\to\mathbb{R}$ be given by $f(x)=\sqrt{x}$ . Claim: $f$ is differentiable.
Proof: Let $c\neq x$ , then

\frac{f(x)-f(c)}{x-c}=\frac{\sqrt{x}-\sqrt{c}}{x-c}\left(\frac{\sqrt{x}+\sqrt{% c}}{\sqrt{x}+\sqrt{c}}\right)=\frac{x-c}{(x-c)(\sqrt{x}+\sqrt{c})}=\frac{1}{% \sqrt{x}+\sqrt{c}}.

(7.5)

Hence $f^{\prime}:(0,\infty)\to\mathbb{R}$ is given by

f^{\prime}(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}\frac{1}{\sqrt{x% }+\sqrt{c}}=\frac{1}{2\sqrt{c}}.

(7.6)

Example 7.3.

Let $f:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=x^{n}$ for some $n\in\mathbb{R}$ . Claim: $f$ is differentiable with derivative $f^{\prime}:\mathbb{R}\to\mathbb{R}$ given by $f(x)=nx^{n-1}$ .
Proof: Let $c\neq x$ , then

\frac{f(x)-f(c)}{x-c}=\frac{x^{n}-c^{n}}{x-c}=\frac{x-c}{x-c}(x^{n-1}+cx^{n-2}% +c^{2}x^{n-3}+\dots+c^{n-2}x+c^{n-1}).

(7.7)

Hence $f^{\prime}(c)$ is given by

f^{\prime}(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}(x^{n-1}+cx^{n-2% }+c^{2}x^{n-3}+\dots+c^{n-2}x+c^{n-1})=nc^{n-1}.

(7.8)

7.1.2 Comparing Differentiability with Continuity

Now that we have defined strictly what a derivative is, we will explore some properties of them. Firstly, we will note that our notion of differentiability is stronger than the definition of continuity.

Theorem 7.1

Let $I\subseteq\mathbb{R}$ be an open interval, let $f:I\to\mathbb{R}$ be a function and let $c\in I$ . Then if $f$ is differentiable at $c$ , then $f$ is continuous at $c$ .

Proof.

Since $f$ is differentiable, $f^{\prime}(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$ . Hence

$\displaystyle\lim_{x\to c}(f(x)-f(c))$	$\displaystyle=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}(x-c)$	(7.9)
	$\displaystyle=\left(\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\right)\left(\lim_{x-c}(% x-c)\right)$	(By theorem LABEL:)
	$\displaystyle=f^{\prime}(c)\cdot 0=0.$	(7.10)

Therefore,

\lim_{x\to c}f(x)=f(c).

(7.11)

So by theorem LABEL:, $f$ is continuous. ∎

Note that the converse of this theorem is not true. Continuity does not imply differentiability (although ).

Example 7.4.

Let $f:(-1,1)\to\mathbb{R}$ be given by $f(x)=\lvert x\rvert$ . Claims:

(i)

$f$ is continuous.
(ii)

$f$ is not differentiable at $x=0$ .

Proof:

(i)

This is true by theorem 6.2.
(ii)

Consider $x\neq 0$ , then

$\frac{f(x)-f(0)}{x-0}=\frac{\lvert x\rvert}{x}=\begin{cases}1\,&x>0\\ -1\,&x<0.\end{cases}$ (7.12)

Thus $f^{\prime}(0)$ is not defined since the limit does not exist at $x=0$ , so $f$ is not differentiable at $x=0$ .

7.1.3 Operations on Differentiable Functions

Now we will prove some properties of derivatives under arithmetic operations similarly to how we have done with every concept we have covered so far. These will be familiar are they are the same rules for derivatives that you will have learned in school.

Theorem 7.2

Let $I\subseteq\mathbb{R}$ , let $f,g:I\to\mathbb{R}$ be two functions, differentiable at $c\in I$ .

(i)

$(f+g):I\to\mathbb{R}$ is differentiable at $c$ and $(f+g)^{\prime}(c)=f^{\prime}(c)+g^{\prime}(c)$ .
(ii)

Given $\lambda\in\mathbb{R}$ , $(\lambda f):I\to\mathbb{R}$ is differentiable at $c$ and $(\lambda f)^{\prime}(c)=\lambda f^{\prime}(c)$ .
(iii)

$(fg):I\to\mathbb{R}$ is differentiable at $c$ and $(fg)^{\prime}(c)=f(c)g^{\prime}(c)+f^{\prime}(c)g(c)$ .
(iv)

If $g(c)\neq 0$ , then $\left(\frac{1}{g}\right):I\to\mathbb{R}$ is differentiable at $c$ and $\left(\frac{1}{g}\right)^{\prime}(c)=\frac{-g^{\prime}(c)}{g(c)^{2}}$ .

Proof.

In all cases below, let $x\in I$ with $x\neq c$ .

(i)

$\displaystyle(f+g)^{\prime}(c)$	$\displaystyle=\lim_{x\to c}\frac{(f+g)(x)-(f+g)(c)}{x-c}$	(7.13)
	$\displaystyle=\lim_{x\to c}\frac{f(x)+g(x)-f(c)-g(c)}{x-c}$	(7.14)
	$\displaystyle=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}+\lim_{x\to c}\frac{g(x)-g(c)}% {x-c}$	(By theorem LABEL:)
	$\displaystyle=f^{\prime}(c)+g^{\prime}(c).$	(7.15)

(ii)

$(\lambda f)^{\prime}(c)=\lim_{x\to c}\frac{(\lambda f)(x)-(\lambda f)(c)}{x-c}% =\lambda\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lambda f^{\prime}(c).$ (7.16)

(iii)

Note that since $f$ and $g$ are differentiable at $c$ , they are continuous at $c$ by theorem 7.1. So $\lim_{x\to c}f(x)=f(c)$ .

$\displaystyle(fg)^{\prime}(c)$	$\displaystyle=\lim_{x\to c}\frac{(fg)(x)-(fg)(c)}{x-c}=\lim_{x\to c}\frac{f(x)% g(x)-f(c)g(c)}{x-c}$	(7.17)
	$\displaystyle=\lim_{x\to c}\frac{f(x)g(x)+f(x)g(c)-f(x)g(c)-f(c)g(c)}{x-c}$	(7.18)
	$\displaystyle=\lim_{x\to c}f(x)\frac{g(x)-g(c)}{x-c}+\lim_{x\to c}g(c)\frac{f(% x)-f(c)}{x-c}$	(7.19)
	$\displaystyle=f(c)g^{\prime}(c)+g(c)f^{\prime}(c).$	(7.20)

(iv)

Since $g(c)\neq 0$ and $g$ is continuous at $c$ , there exists $\delta>0$ so that $x\in(c-\delta,c+\delta)\implies g(x)\neq 0$ .

$\displaystyle\left(\frac{1}{g}\right)^{\prime}(c)=\lim_{x\to c}\frac{\left(% \frac{1}{g}\right)(x)-\left(\frac{1}{g}\right)(c)}{x-c}$	$\displaystyle=\lim_{x\to c}\left(\frac{1}{x-c}\right)\left(\frac{1}{g(x)}-% \frac{1}{g(c)}\right)$	(7.21)
	$\displaystyle=\lim_{x\to c}\left(\frac{1}{x-c}\right)\left(\frac{g(c)-g(x)}{g(% x)g(c)}\right)$	(7.22)
	$\displaystyle=\lim_{x\to c}\left(\frac{-1}{g(x)g(c)}\right)\frac{g(x)-g(c)}{x-c}$	(7.23)
	$\displaystyle=\frac{-g^{\prime}(c)}{g(c)^{2}}.$	(7.24)

∎

Part (iii) of this theorem is the well-known product rule for derivatives. We can prove the quotient rule as a corollary.

Corollary 7.2.1 (Quotient Rule)

If $g(c)\neq 0$ , $\left(\frac{f}{g}\right):I\to\mathbb{R}$ is differentiable at $c$ and ${\left(\frac{f}{g}\right)^{\prime}(c)=\frac{g(c)f^{\prime}(c)-g^{\prime}(c)f(c% )}{g(c)^{2}}}$ .

Proof.

Since $\frac{f}{g}=f\cdot\frac{1}{g}$ ,

$\displaystyle\left(f\cdot\frac{1}{g}\right)^{\prime}(c)$	$\displaystyle=\frac{f^{\prime}(c)}{g(c)}+f(c)\left(\frac{1}{g}\right)^{\prime}% (c)$	(By theorem 7.2 (iii))
	$\displaystyle=\frac{f^{\prime}(c)}{g(c)}-\frac{f(c)g^{\prime}(c)}{g(c)^{2}}$	(By theorem 7.2 (iv))
	$\displaystyle=\frac{g(c)f^{\prime}(c)-g^{\prime}(c)f(c)}{g(c)^{2}}.$	(7.25)

∎

Other corollaries of this theorem are that all polynomials are differentiable, and all rational functions are differentiable except where the denominator is zero.

Theorem 7.3 (Chain Rule)

Let $I,J\subseteq\mathbb{R}$ be intervals, $f:I\to\mathbb{R}$ a function differentiable at some $c\in I$ with $f(I)\subseteq J$ , $g:J\to\mathbb{R}$ a function differentiable at $f(c)\in J$ . Then $(g\circ f):I\to\mathbb{R}$ is differentiable at $c$ and $(g\circ f)^{\prime}(c)=g^{\prime}(f(c))f^{\prime}(c)$ .