NoteNextra-origin/content/Math4121/Math4121_L2.md

Math4121 Lecture 2

Chapter 5: Differentiation

Continue on Differentiation

Theorem 5.5: Chain Rule

Suppose

1. f:[a,b]\to \mathbb{R} is continuous on [a,b] (or some neighborhood of x)
2. f'(x) exists at some point x\in (a,b) (f is differentiable at x)
3. g is defined on an interval [c,d] containing the range of f, (f([a,b])\subset [c,d])
4. 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

Example 5.6

(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.

Mean Value Theorem

Definition 5.7: Local Extrema

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.

Theorem 5.8

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