diff --git a/pages/Math4121/Math4121_L2.md b/pages/Math4121/Math4121_L2.md index 2707d3e..c2002c8 100644 --- a/pages/Math4121/Math4121_L2.md +++ b/pages/Math4121/Math4121_L2.md @@ -1 +1,138 @@ -# Lecture 2 \ No newline at end of file +# 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 $h'(x)=g'(y)f'(x)$. + +EOP + +#### 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