5.3 Differentiation laws

When studying computational calculus, you will likely have learned useful rules for differentiating sums, products, quotients and compositions of functions. One example is the product rule $(fg)^{\prime}=f^{\prime}g+fg^{\prime}$ and another is the chain rule $(g\circ f)^{\prime}=(g^{\prime}\circ f)f^{\prime}$ . However, can you explain where these rules come from and why they hold? In this section, we revisit these computational tools and show that they all inexorably follow from the notion of a derivative introduced in Definition 5.2.

Theorem 5.24.

Let $I\subseteq\mathbb{R}$ be an open interval, $f$ , $g\colon I\to\mathbb{R}$ be differentiable at $a\in I$ .

1

Linearity. The function $\lambda f$ for $\lambda\in\mathbb{R}$ and $f+g\colon I\to\mathbb{R}$ are differentiable at $a$ , with

$(\lambda f)^{\prime}(a)=\lambda f^{\prime}(a)\qquad\text{and}\qquad(f+g)^{% \prime}(a)=f^{\prime}(a)+g^{\prime}(a).$
2

Leibniz/product rule. The function $f\cdot g\colon I\to\mathbb{R}$ is differentiable at $a$ and

$(f\cdot g)^{\prime}(a)=f^{\prime}(a)g(a)+f(a)g^{\prime}(a).$
3

Quotient rule. Provided $g(x)\neq 0$ for all $x\in I$ , the function $(f/g)\colon I\to\mathbb{R}$ is differentiable at $a$ and

$\big{(}f/g\big{)}^{\prime}(a)=\frac{f^{\prime}(a)g(a)-f(a)g^{\prime}(a)}{g(a)^% {2}}.$

Proof.

We only prove part (b); the remaining parts are left as exercises (see Exercise 5.25 and Worksheet 9).

By Lemma 5.21, there exist functions $F$ , $G\colon I\to\mathbb{R}$ that are continuous at $a$ with $F(a)=f^{\prime}(a)$ and $G(a)=g^{\prime}(a)$ and satisfy

(5.10) (5.10)

f(x)=F(x)(x-a)+f(a)\qquad\text{and}\qquad g(x)=G(x)(x-a)+g(a)\qquad\text{for % all $x\in I$.}

By substituting the formulæ from (5.10) into $f(x)g(x)$ and expanding out the resulting product,

f(x)g(x)=H(x)(x-a)+f(a)g(a)

where

(5.11) (5.11)

H(x):=F(x)g(a)+f(a)G(x)+F(x)G(x)(x-a)\quad\text{for all $x\in I$.}

By the continuity laws from Corollary 4.51, the function $H\colon I\to\mathbb{R}$ is continuous at $a$ . It therefore follows that

(fg)^{\prime}(a)=\lim_{x\to a}\frac{f(x)g(x)-f(a)g(a)}{x-a}=\lim_{x\to a}H(x)=% H(a).

However, using the definition of $H$ from (5.11) and the fact that $F(a)=f^{\prime}(a)$ and $G(a)=g^{\prime}(a)$ , we see that $H(a)=f^{\prime}(a)g(a)+f(a)g^{\prime}(a)$ . Combining these observations concludes the proof. ∎

Exercise 5.25.

Prove Theorem 5.24 1) by arguing from the definition of the derivative.

Example 5.26 (Polynomials).

Let $p\colon\mathbb{R}\to\mathbb{R}$ be a polynomial function, so that there exists some $d\in\mathbb{N}$ and coefficients $c_{0}$ , $c_{1}$ , $\dots$ , $c_{d}\in\mathbb{R}$ such that

p(x):=c_{d}x^{d}+c_{d-1}x^{d-1}+\cdots+c_{1}x+c_{0}\qquad\text{for all $x\in% \mathbb{R}$.}

By Example 5.11 and the linearity of the derivative, it follows that $p$ is differentiable and

p^{\prime}(x):=c_{d}dx^{d-1}+c_{d-1}(d-2)x^{d-2}+\cdots+c_{1}\qquad\text{for % all $x\in\mathbb{R}$.}

Example 5.27 (More trigonometric functions).

The function $\tan\colon(-\pi/2,\pi/2)\to\mathbb{R}$ is differentiable with $\tan^{\prime}(x)=1/\cos^{2}(x)$ for all $x\in(-\pi/2,\pi/2)$ . To see this, we apply the quotient rule to the definition of $\tan:=\sin/\cos$ . Since both $\sin$ and $\cos$ are differentiable and $\cos$ is non-vanishing on $(-\pi/2,\pi/2)$ , this tells us that $\tan$ is differentiable and

\tan^{\prime}(x)=\frac{\sin^{\prime}(x)\cos(x)-\sin(x)\cos^{\prime}(x)}{\cos^{% 2}(x)}=\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos^{2}(x)}=\frac{1}{\cos^{2}(x)}

for all $x\in(-\pi/2,\pi/2)$ , where we have used Example 5.13 and the identity $\sin^{2}(x)+\cos^{2}(x)=1$ .

Theorem 5.28 (Chain Rule).

Let $I$ , $J\subseteq\mathbb{R}$ be open intervals, $f\colon I\to J$ be differentiable at $a\in I$ and $g\colon J\to\mathbb{R}$ be differentiable at $f(a)\in J$ . Then $g\circ f$ is differentiable at $a$ and

(g\circ f)^{\prime}(a)=g^{\prime}(f(a))f^{\prime}(a).

Proof.

By Lemma 5.21, there exists a function $F\colon I\to\mathbb{R}$ which is continuous at $a$ with $F(a)=f^{\prime}(a)$ and a function $G\colon J\to\mathbb{R}$ which is continuous at $f(a)$ with $G(f(a))=g^{\prime}(f(a))$ which satisfy

(5.12) (5.12)

f(x)=F(x)(x-a)+f(a)\quad\text{and}\qquad g(y)=G(y)(y-f(a))+g(f(a))\qquad\text{% for all $x\in I$, $y\in J$.}

By applying the formulæ (5.12) to $(g\circ f)(x)=g(f(x))$ with $y=f(x)$ , we have

$\displaystyle(g\circ f)(x)$	$\displaystyle=G(y)(y-f(a)+g(f(a))$
	$\displaystyle=G(f(x))(f(x)-f(a))+g(f(a))$
	$\displaystyle=H(x)(x-a)+(g\circ f)(a),$

where

(5.13) (5.13)

H(x):=G(f(x))F(x)\qquad\text{for all $x\in I$.}

By the continuity laws, $H\colon I\to\mathbb{R}$ is continuous at $a$ . It therefore follows that

(g\circ f)^{\prime}(a)=\lim_{x\to a}\frac{(g\circ f)(x)-(g\circ f)(a)}{x-a}=% \lim_{x\to a}H(x)=H(a)

However, using the definition of $H$ from (5.13) and the fact that $F(a)=f^{\prime}(a)$ and $G(f(a))=g^{\prime}(f(a))$ , we see that $H(a)=f^{\prime}(a)g^{\prime}(f(a))$ . Combining these observations concludes the proof. ∎

Figure 5.8: The function

f\colon\mathbb{R}\to\mathbb{R}

defined in (5.14) is differentiable, but

f^{\prime}

is not continuous at

0

. You can zoom in using an interactive version of this graph at https://www.desmos.com/calculator/or4d5tqvtz.

We can use the chain rule and the other laws for differentiation to analyse the following example, which exhibits some interesting behaviour.

Example 5.29.

Consider function $f\colon\mathbb{R}\to\mathbb{R}$ given by

(5.14) (5.14)

f(x):=\begin{cases}x^{2}\sin(1/x)&\text{if $x\neq 0$,}\\ 0&\text{if $x=0$;}\end{cases}

see Figure 5.8. Then $f$ is differentiable but its derivative $f^{\prime}$ is not continuous.

Proof.

By the rules for differentiation, the function $f$ is differentiable¹¹ 1 Strictly speaking, we only defined differentiability for functions defined on an open interval $I$ and $\mathbb{R}\setminus\{0\}$ is not an interval. Here, when we say $f$ is differentiable on $\mathbb{R}\setminus\{0\}=(-\infty,0)\cup(0,\infty)$ , we mean that it is differentiable on both of the constituent open intervals $(-\infty,0)$ and $(0,\infty)$ . Similar remarks apply to other domains which can be written as a union of open intervals. on $\mathbb{R}\setminus\{0\}$ with

(5.15) (5.15)

f^{\prime}(x)=2x\sin(1/x)-\cos(1/x)\qquad\text{for all $x\in\mathbb{R}% \setminus\{0\}$.}

It remains to check whether it is differentiable at $0$ . The difference quotient is given by

\frac{f(0+h)-f(0)}{h}=\frac{h^{2}\sin(1/h)-0}{h}=h\sin(1/h)\qquad\text{for $h% \in\mathbb{R}\setminus\{0\}$.}

Since $|\sin(1/h)|\leq 1$ , it follows from the squeeze theorem that $f$ is indeed differentiable at $0$ with

f^{\prime}(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=0.

Finally, in light of (5.15), the function $f^{\prime}$ is not continuous at $0$ . Indeed, similar to Example 4.36, the rapid oscillation of $\cos(1/x)$ means that $\lim_{x\to 0}f^{\prime}(x)$ does not exist. ∎