Math4121 Lecture 2

Chapter 5: Differentiation

Suppose

$f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ (or some neighborhood of $x$ )
$f'(x)$ exists at some point $x\in (a,b)$ ( $f$ is differentiable at $x$ )
$g$ is defined on an interval $[c,d]$ containing the range of $f$ , ( $f([a,b])\subset [c,d]$ )
$g$ is differentiable at the point $f(x)$

Let $h=g\circ f$ ( $h=g(f(x))$ ) where $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ . Then $h$ is differentiable at $x$ and

h'(x) = g'(f(x))f'(x)

Proof:

Let $y=f(x)$ and $u(t)=\frac{f(t)-f(x)}{t-x}-f'(x)$ for $t\neq x,t\in [a,b]$ , $v(s)=\frac{g(s)-g(y)}{s-y}-g'(y)$ for $s\neq y,s\in [c,d]$ .

Notice that $u(t)\to 0$ as $t\to x$ and $v(s)\to 0$ as $s\to y$ .

Pick $s=f(t)$ for $t\in [a,b]$ so that $s\to y$ as $t\to x$ . Then

\begin{aligned} h(t)-h(x) &= g(f(t))-g(f(x)) \\ &= g(t)-g(y) \\ &= (s-y)(g'(y)+v(s)) \\ &= (f(t)-f(x))(g'(y)+v(s)) \\ &= (t-x)(f'(x)+u(t))(g'(y)+v(s)) \\ \end{aligned}

So $h'(x)=\frac{h(t)-h(x)}{t-x}=(f'(x)+u(t))(g'(y)+v(s))$ . Since $u(t)\to 0$ and $v(s)\to 0$ as $t\to x$ and $s\to y$ , we have $h'(x)=g'(y)f'(x)$ .

QED

(a) Let $f(x)=\begin{cases} x\sin\frac{1}{x} & x\neq 0 \\ 0 & x=0 \end{cases}$

For $x\neq 0$ ,

\begin{aligned} f'(x) &= 1\cdot\sin\frac{1}{x}+x\cos\frac{1}{x}\cdot\frac{-1}{x^2} \\ &= \sin\frac{1}{x}-\frac{\cos\frac{1}{x}}{x} \end{aligned}

For $x=0$ ,

\begin{aligned} f'(0) &= \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} \\ &= \lim_{x\to 0}\frac{x\sin\frac{1}{x}}{x} \\ &= \lim_{x\to 0}\sin\frac{1}{x} \end{aligned}

This limit does not exist, so $f$ is not differentiable at $x=0$ .

(b) Let $f(x)=\begin{cases} x^2 \sin\frac{1}{x} & x\neq 0 \\ 0 & x=0 \end{cases}$

For $x\neq 0$ ,

\begin{aligned} f'(x) &= 2x\sin\frac{1}{x}+x^2\cos\frac{1}{x}\cdot\frac{-1}{x^2} \\ &= 2x\sin\frac{1}{x}-\cos\frac{1}{x} \end{aligned}

For $x=0$ ,

\begin{aligned} f'(0) &= \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} \\ &= \lim_{x\to 0}\frac{x^2\sin\frac{1}{x}}{x} \\ &= \lim_{x\to 0}x\sin\frac{1}{x}\\ &= 0 \end{aligned}

So $f'(x)=\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} & x\neq 0 \\ 0 & x=0 \end{cases}$ .

Notice that $f'(x)$ is not continuous at $x=0$ since $\lim_{x\to 0}f'(x)$ is undefined.

Let $f:[a,b]\to \mathbb{R}$ . We say that $f$ has a local maximum (or minimum) at $x\in [a,b]$ if there exists some $\delta>0$ such that

f(x)\geq f(t) \text{ for all }|x-t|<\delta

for local maximum, and

f(x)\leq f(t) \text{ for all }|x-t|<\delta

for local minimum.

If $f:[a,b]\to \mathbb{R}$ has a local maximum (or minimum) at $x\in (a,b)$ and $f$ is differentiable at $x$ , then $f'(x)=0$ .

Proof:

We can find $\delta>0$ such that $a<x-\delta<x<x+\delta<b$ .

And for all $x-\delta<t<x+\delta$ ,

If $x-\delta<t<x$ , then $f(x)\geq f(t)$ so $\frac{f(t)-f(x)}{t-x}\leq 0$ .

If $x<t<x+\delta$ , then $f(x)\geq f(t)$ so $\frac{f(t)-f(x)}{t-x}\geq 0$ .

So $\lim_{t\to x}\frac{f(t)-f(x)}{t-x}=0$ .

QED