MATH 137

0 | Pre-Calculus Review

0.1 | Real-Valued Functions

Function:

• Vertical Line Test
• Let X and Y be sets. A function f is a mapping that assigns to each $x\in X$ exactly one $y=f(x)\in Y.$ We use the notation

$$f:X\to Y,x\to f(x)$$

Domain and Range

• Domain: List of possible inputs
• Range: List of possible outputs

$$range(f)=\{f(x):x\in D\}$$

Examples

1. $y=x$
   1. $D=\mathbb{R},R=\mathbb{R}$
2. $y=x^2$
   1. $D=\mathbb{R},R=[0,\infty )$
3. $y=\frac{1}{x}$
   1. $D=\mathbb{R}-0,R=\mathbb{R}-0$

Odd and Even Functions

• $f$ is called even if $f(x)=f(-x)$ for all $x\in D$
• $g$ is called odd if $g(x)=-g(-x)$ for all $x\in D$
• Suppose $f$ is a function, and suppose that $x\in D$ so that $f(x)=0$. Then we call x a root of f

Function Composition

$$g\circ f=g(f(x))$$

Examples

1. $6x+3$
2. $6x+9$
3. $4x+9$
4. $9x$

Functional Inverses (only works with bijective/passes vertical line test)

$$f^{-1}(f(x))=x$$

Given $g(x)=\frac{1}{2}x+7$ find $g^{-1}(x)$

$y=\frac{1}{2}x+7$ $y-7=\frac{1}{2}x$ $x=2(y-7)$ $x=2y-14$ $g^{-1}(x)=2x-14$

Examples

Find the Domain

1. $x\in \mathbb{R}$
2. $x\in \mathbb{R}-\{2,-2\}$

Even or Odd

1. Neither
2. Odd
3. Even

Find the Inverse

1. $f^{-1}(x)=\sqrt{\frac{4x-13}{13}}$
2. $x=3y^2-2y+5,f^{-1}(5)=0$

0.2 | Polynomials

Def: Polynomials are functions of the form

$$f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_0$$

Number of Roots

• Fundamental Theorem of Arithmetic: Polynomials with degree n have n complex roots

Ex: Long Division

$$\frac{6x^3-8x+5}{2x-4}=(3x^2+6x+8)(2x-4)+37$$

Ex: When $x^3-4x^2+mx-2$ is divided by $x-1$, the remainder is -7. What is m?

$$m=-2$$

1 | Sequence Limits

1.1 | Absolute Values

Triangle Inequality 1

Ex: Triangle Inequality 1: > For all $x,y,z\in \mathbb{R},$

$$|x-y|\le |x-z|+|z-y|$$

Proof: Assume WLOG that $x\le y$ Case 1: $z<x$ Case 2: $x\le z\le y$ Case 3: $y<z$

Triangle Inequality 2

Triangle Inequality 2: For all $x,y\in \mathbb{R}$

$$x+y\le |x|+|y|$$

Proof:

$$\begin{array}{rl}|x+y|& \le |x-0|+|0-y|\\ & =|x|+|y|\end{array}$$

Absolute Value Problems

Ex 1: Single Bound

$$|x-2|<3$$

Simplifying: $-3<x-2<3$ $-3+2<x<3+2$ $-1<x<5$

Ex 4: Double Bound

$$2\le |x-4|<4$$ $$x\in (0.2]\cup [6,8)$$

Ex 5: Double Absolute

$$|x-3|-|x-4|<x$$

Ex 3: Double Absolute

$$|7x-3|\ge |3x+7|$$ $$x\in \left(-\infty ,-\frac{2}{5}\right)\cup \left(\frac{5}{2},\infty \right)$$

1.2 | Sequences and Limits

Uniqueness of Sequence Limit

(Uniqueness of Sequence Limit) Let $\{a_n\}$ be a sequence. If $\{a_n\}$ has a limit L, then the limit L is unique.

1.3 | Rules for Limits

Epsilon-N

Epsilon-Delta definition of a limit

Let $\{a_n\}$ be a sequence and $L\in \mathbb{R}$. We say that L is the limit of $\{a_n\}$ if for all $ϵ>0$ there exists a real number N such that if $n>N$, then

$$|a_n-L|<ϵ$$

If such an L exists, we say that $\{a_n\}$ CONVERGES to L and write

$$\underset{n\to \infty }{\lim }{a}_n=L\text{ OR }{a}_n\to L$$

If no such L exists, we say that $\{a_n\}$ DIVERGES

Ex: Prove that $a_n=\frac{1}{n^2}$ converges to 0

Proof: Let $ϵ>0$ be given. We determine N so that $|a_n-L|<ϵ$ by solving

$$a_n-L<ϵ$$ $$|\frac{1}{n^2}|<ϵ$$ $$\frac{1}{n^2}<ϵ$$ $$n^2>\frac{1}{ϵ}$$ $$n>\frac{1}{\sqrt{ϵ}}$$

Thus, for every $ϵ>0$, let $N=\frac{1}{\sqrt{ϵ}}$. Then, for all $n>N$,

$$|a_n-L|=\frac{1}{n^2}<\frac{1}{{\left(\frac{1}{\sqrt{ϵ}}\right)}^2}$$

More Epsilon-N

Epsilon-Delta: every function - its limit will always be less than the smallest real number

Show that ${\lim }_{n\to \infty }\frac{3n+2}{4n+3}=\frac{3}{4}$ for all arbitrary $ϵ$

USING $ϵ=\frac{1}{1000}$

$$a_n=\frac{3n+2}{4n+3},L=\frac{3}{4}$$ $$|a_n-L|=|\frac{(3n+2)}{4n+3}-\frac{3}{4}|=|-\frac{1}{16n+12}|=\frac{1}{16n+12}<\frac{1}{1000}$$ $$\frac{\frac{1}{16n+12}<1}{1000}⟹16n+12>1000⟹n>61.75$$

AKA: the bound is too big for all $n>61.75$. The bound should approach infinity.

USING $ϵ\to 0$

$$a_n=\frac{3n+2}{4n+3},L=\frac{3}{4},N=\frac{1}{16}\left(\frac{1}{ϵ-12}\right)$$ $$|a_n-L|=\frac{1}{16n+12}<\frac{1}{16\left(\frac{1}{16}\left(\frac{1}{ϵ}-12\right)\right)+12}$$ $$|a_n-L=|\frac{3n+2}{4n+3}-\frac{3}{4}|=|-\frac{1}{16n+12}|=\frac{1}{16n+12}<ϵ$$ $$\frac{1}{16n+12}<ϵ⟹16n+12>\frac{1}{ϵ}⟹n>\frac{1}{16}\left(\frac{1}{ϵ}-12\right)=N$$

1.4 | Squeeze Theorem

Squeeze Theorem

(Squeeze Theorem for Sequences) If $a_n\le b_n\le c_n$ for all $n>M$ (for some $M\in \mathbb{N}$) and ${\lim }_{n\to \infty }{a}_n=L={\lim }_{n\to \infty }$, then ${\lim }_{n\to \infty }{b}_n=L$ as well

Use the Squeeze Theorem to prove convergence:

$$a_n=\frac{(-1{)}^n}{n^2+1}$$

Solution:

$$\frac{-1}{n^2+1}\le \frac{(-1{)}^n}{n^2+1}\le \frac{1}{n^2+1}$$ $$\underset{n\to \infty }{\lim }\frac{-1}{n^2+1}=0=\underset{n\to \infty }{\lim }\frac{1}{n^2+1}$$

Hence, ${\lim }_{n\to \infty }\frac{(-1{)}^n}{n^2+1}=0$ by the squeeze theorem.

1.5 | Recursive Sequences

Explicit vs Recursive

Explicit Sequence

$${\left\{\frac{1}{n+1}\right\}}_{n=1}^{\infty }$$

Recursive Sequences

$$F_1=1,F_2=1,F_n=F_{n-1}+F_{n-2}n\ge 3$$

GLB and LUB

Let $S\subset \mathbb{R}$. Then $\beta$ is called the GREATEST LOWER BOUND of $S$ if :

1. $\beta$ is a lower bound, and
2. $\beta$ is the largest lower bound, that is, if ${\beta }^{\prime }$ is another lower bound of S, then ${\beta }^{\prime }\le \beta$ Greatest Lower Bound of S = $glb(s)$ = infimum = $inf(s)$

Greatest Lower Bound $(glb)$ and Least Upper Bound $(lub)$

Ex: $S=[0,1)$ is bounded

• $glb(S)=0$
• $lub(S)=1$
• NOTE: $glb$ and $lub$ might be in S, but do not have to be

Ex: $S=\{0\}\cup [1,2]\cup \{3\}$

• $lub(S)=3$
• $glb(s)=0$

Ex: $S=x\in \mathbb{Q}:x^2<2$

• $lub(S)=\sqrt{2}$
• $glb(s)=-\sqrt{2}$

Monotone Convergence Theorem:

MONOTONE = Strictly increasing or decreasing

(Theorem 1.5.7 in notes) Let $\{a_n\}$ be an increasing sequence

1. If $\{a_n\}$ is bounded above, then $\{a_n\}$ converges to $L=lub(\{a_n\})$.
2. If $\{a_n\}$ is not bounded above, then$\{a_n\}$ diverges to $\infty$ Similarly, let $\{b_n\}$ be a decreasing sequence
3. If $\{b_n\}$is bounded below, then $\{b_n\}$ converges to $L=glb(\{b_n\})$
4. If $\{b_n\}$is not bounded below, then $\{b_n\}$ diverges to $-\infty$

Ex: We have $\{a_n{\}}_{n=1}^{\infty },a_1=1,a_{n+1}=\frac{1+2a_n}{5}$ If a converges, then ${\lim }_{n\to \infty }{a}_n=\frac{1}{5}$

We will show that the sequence converges using MCT To do this, we use induction to show that it is decreasing and bounded below. Base case: $n=1$. Then $a_n=1>0$ Induction Step: Assume $a_k>0$ for some $k\ge 1$. We need to show, using $a_k>0$, that $a_{k+1}=0$. Since $a_k>0$, $2a_k>2(0)=0$, so $2a_k+1>1$. Dividing by 5 gives $\frac{1+2a_k}{5}>\frac{1}{5}=0$ Therefore, $a_{k+1}>0$ By induction, $a_n>0$ for all n.

Now, we prove that $a_n$ is decreasing, or that $a_n>a_{n+1}$ for all n. Base case: $n=1,a_n=a_1=1$ $a_{n+1}=a_2=\frac{1+2a_1}{5}=\frac{1}{5}$ $\frac{3}{5}$ so $a_n>a_{n+1}$ when $n=1$ Induction step: Assume $a_k>a_{k+1}$ for some $k\ge 1$ Starting with $a_k>a_{k+1}$, we have that $2a_k>2a_{k+1}$ $1+2a_k>1+2a_{k+1}$ $\frac{1+2a_k}{5}>\frac{1+2a_{k+1}}{5}$ $a_{k+1}>a_{k+2}$ Therefore, $a_n>a_{n+1}$ for all $n\ge 1$ by induction. Since $\{a_n{\}}_{n=1}^{\infty }$ is decreasing and bounded below, it converges by the MCT.

Let $a_1=\sqrt{2}$ and $a_{n+1}=\sqrt{2+9n}$ for all $n\ge 1$. Prove that $\{a_n{\}}_{n=1}^{\infty }$

2 | Function Limits and Continuity

2.1 | Introduction to Function Limits

Epsilon-Delta

Example: Consider

$$f(x)=\frac{(x+1)(x+2)}{x+1}$$

• -1 is not in the domain
• ${\lim }_{x\to a}$ has nothing to do with $f(a)$. In fact, it can exist even when $f(a)$ is undefined.

Informal definition of a limit: The limit as $x$ approaches $a$ of $f(x)$ is $L$ if $f(x)$ is very close to $L$ when x is very close to $a$

(Definition) Formal Limit Definition Let $f(x)$ be a function and $a\in \mathbb{R}$. We say that the limit as $f(x)$ approaches a from the left is L and write

$$\underset{x\to a^{-}}{\lim }f(x)=L$$

If for all $ϵ$ there exists $\delta >0$ such that

$$0<|x-a|<\delta ⟹|f(x)-L|<ϵ$$

Epsilon-Delta Examples

Example: Show that ${\lim }_{x\to 3}3x+1=10$

Aside: $|f(x)-L|<ϵ$ $|(3x+1)-10|<ϵ$ $|3x+1-10|=|3x-9|=3|x-3|<ϵ$ $|x-3|<\frac{ϵ}{3}$

Proof: Let $ϵ>0$ be given and choose $\delta =\frac{ϵ}{3}$. If $0<|x-3|<5$ then $|3x+1-10|=|3x-9|=3|x-3|<ϵ=3\delta$ But $\delta =\frac{ϵ}{3}$ so $|3x+1-10|<3\times \frac{5}{3}=ϵ$ So $0<|x-3|<\delta$ implies $|3x+1-10|<ϵ$, so ${\lim }_{x\to 3}3x+1=10$

Example: Prove that ${\lim }_{x\to 7}{x}^2+1=50$

Rough Work: By making $|x-7|$ small we want to make $|(x^2+1)-50|$ small.

$$\begin{array}{rl}|x^2+1-50|& =|x^2-49|\\ & =|x-7||x+7|\end{array}$$

Proof must ensure that $\delta \le 1$

$$\begin{array}{rl}|x+7|=|x-7+14|& \le |x-7|+|14|\\ & <ϵ+14\le 15\end{array}$$

$|x-7||x+7|<\delta \cdot 15=ϵ$ $⟹\delta =\frac{ϵ}{15}$

Proof: Let $ϵ>0$ be given. Choose $\delta =min(1,\frac{ϵ}{15})$. Thus $ϵ\le 1$ and $\delta \le \frac{ϵ}{15}$. Assume $0<|x-7|<\delta$ Note that

$$|x+7|=|x-7+14|\le |x-7|+|14|<\delta +14\le 1+14=15⟹|x+7|<15$$

Now show that $x^2+1-50<ϵ$

$$\begin{array}{rl}|x^2+1-50|& =|x^2-49|\\ & =|x-7||x+7|\\ & <15\cdot \delta \\ & \le 15\cdot \frac{\delta }{15}=ϵ\end{array}$$

Ex: Show that ${\lim }_{x\to 0}\frac{|x|}{x}$ does not exist.

Proof; Assume towards a contradiction that ${\lim }_{x\to \infty }\frac{|x|}{x}=L$ for some $L\in \mathbb{R}$ Consider $ϵ=\frac{1}{2}$. Since L exists, there is $\delta >0$ such that

$$0<|x-0|<\delta ⟹|\frac{|x|}{x}-L|<\frac{1}{2}$$

Let $x_1=-\frac{\delta }{2},x_2=\frac{\delta }{2}$ $|x_1-0|=|-\frac{\delta }{2}|=|-\frac{\delta |}{2}=\frac{\delta }{2}<\delta$ So $|\frac{|x_1|}{x_1}|<\frac{1}{2}$ $x_1<0$, so this is equivalent to $|-1-L|<\frac{1}{2}$ Similarly, $|x_2-0|=|\frac{\delta }{2}|=\frac{\delta }{2}<\delta$ So $|\frac{|x_2|}{x_2}-L|=|1-L|<\frac{1}{2}$ since $x_2>0$

$$\begin{array}{rl}2& =|1-(-1)|\\ & =|(1-l)+(L-(-1))|\\ & \le |1-L|+|L-(-1)|\\ & =|1-L|+|-1-L|\\ & <\frac{1}{2}+\frac{1}{2}=1\\ 2& <1\end{array}$$

Contradiction!

2.2 | Sequential Characterization of Limits

2.3 | One-Sided Limits

Example: show that ${\lim }_{x\to \infty }\frac{|x|}{x}$ does not exist.

PROOF: It suffices to prove that the limits from both directions are not equal. Show ${\lim }_{x\to {\infty }^{+}}\frac{|x|}{x}=1$ Let $ϵ>0$ and choose $\delta =7$ Then $0<x-0<\delta ⟹$ 0<x<7 so $|x|=x$

$$|\frac{|x|}{x}-1|=|\frac{x}{x}-1|=|1-1|=0<ϵ$$

Similarly, ${\lim }_{x\to {\infty }^{-}}f(x)=-1$ so the one sided limits are not equal.

2.4 | Fundamental Trig Limit

(Theorem) Fundamental Trig Limit)

$$\underset{x\to \infty }{\lim }\frac{sin(x)}{x}=1$$

Example: Show that ${\lim }_{x\to {\infty }^{+}}x\sin (\frac{1}{x})=0$

$-1\le \sin (\frac{1}{x})$ for all $x\ne 0$ If $x=0$ then $-x\le x\sin (\frac{1}{x})=\le x$ ${\lim }_{x\to {\infty }^{+}}={\lim }_{x\to {\infty }^{+}}-x=0$ So ${\lim }_x\to {\infty }^{+}x\sin (\frac{1}{x})=0$ Similarly, ${\lim }_x\to {\infty }^{-}x\sin (\frac{1}{x})=0$ So ${\lim }_{x\to \infty }x\sin (\frac{1}{x})=0$

Example; Compute ${\lim }_{x\to \infty }\frac{sin(4x)}{sin(5x)}$

$={\lim }_{x\to \infty }\frac{4\cdot 5x}{5\cdot 4x}\cdot \frac{sin(4x)}{sin(5x)}$ $=\frac{4}{5}{\lim }_{x\to \infty }\frac{5x}{sin(5x)}\cdot \frac{sin(4x)}{4x}$ $=\frac{4}{5}{\lim }_{x\to \infty }\frac{\frac{\frac{sin(x)}{x}}{sin(5x)}}{5x}$ $=\frac{4}{5}\cdot \frac{1}{1}$ $=\frac{4}{5}$

2.5 | Limits at Infinity and Horizontal Asymptotes

(Definition) Limits as x approaches infinity Let $f(x)$ be a function. We say that $f(x)$ approaches L as x approaches $\infty$ and write ${\lim }_{x\to \infty }f(x)=L$ if for $ϵ>0$ there exists $N\in \mathbb{R}$ such that if $x>N$, then $|f(x)-L|<ϵ$

(Definition) Horizontal Asymptotes Let $f(x)$ be a function and $L\in \mathbb{R}$. We say that the line with equation $y=L$ is a horizontal asymptote of $f(x)$ if wither ${\lim }_{x\to \infty }f(x)=L$ or ${\lim }_{x\to -\infty }f(x)=L$

2.6 | Fundamental Log Limit

(Definition) Fundamental Log Limit

$$\underset{x\to \infty }{\lim }\frac{ln(x)}{x}=0$$

2.7 | Infinite Limits and Vertical Asymptotes

(Definition) Limits to Infinity We say that $f(x)\to \infty$ as $x\to a$ from the right if for all $M>0$ there exists $\delta >0$ such that if $0<x-a<\delta$ ${\lim }_{x\to \infty }=\infty$ then $f(x)>M$ ${\lim }_{x\to a^{-}}f(x)=\infty ,{\lim }_{x\to a^{+}}f(x)=-\infty$ and ${\lim }_{x\to {\infty }^{-}}f(x)=\infty$ are all defined

Similarly, if ${\lim }_{x\to a^{+}}f(x)=\pm \infty$ or ${\lim }_{a\to a^{-}}f(x)=\pm \infty$, we say that $f(x)$ has a VERTICAL ASYMPTOTE at $x=a$.

Example: ${\lim }_{x\to 2^{-}}\frac{x+1}{x-2}=\frac{3}{0}$ So the limit is $\pm \infty$

2.8 | Continuity

Fact about continuity: $f(x)$ is continuous at $x=a$ if and only if ${\lim }_{h\to 0}f(a+h)=f(a)$

(Theorems 2.8.7, 2.8.8, 2.8.9) The following functions are continuous everywhere in their domain:

1. Polynomials
   1. Proof: We can evaluate by substitution: ${\lim }_{x\to a}P(x)=P(a)$
2. $sin(x),cos(x)$
   1. Proof: ${\lim }_{h\to 0}f(a+h)={\lim }_{h\to 0}sin(a+h)={\lim }_{h\to 0}(sin(a)cos(h)+sin(h)cos(a)$
3. $e^x$
   1. ${\lim }_{h\to 0}{e}^{a+h}={\lim }_{h\to 0}{e}^a\cdot e^h=e^a{\lim }_{h\to \infty }{e}^h$
   2. Proof: ${\lim }_{h\to \infty }{e}^h=1=e^0$
4. $ln(x)$

(Theorem) Let $f(x),g(x)$ be continuous at $x=a$. The following are also continuous at $x=a$.

1. $f(x)+g(x)$
2. $f(x)g(x)$
3. $\frac{f(x)}{g(x)}$
4. $cf(x)$

(Theorem): If $f(x)$ is continuous at $x=a$, so it it’s inverse.

(Theorem): Suppose $f(x),g(x)$ are continuous at $x=a$, then $g\circ f$ is continuous at $x=a$

(Proof) Sequential Characterization

$f(x)$ is continuous at $x=a$, ${\lim }_{n\to \infty }f(xn)=f(a)$ by sequential characterization Now $\{f(x_n){\}}_{n=1}^{\infty }$ is a sequence converging to $f(a)$ By continuity of $f(x)$ at $x=f(a)$ and the sequential characterization ${\lim }_{n\to \infty }g(f(x_n))=g(f(n))$. By the sequential characterization, $g\circ f$ is continuous at $x=a$

(Definition) We say $f(x)$ is continuous on $[a,b]$ if ${\lim }_{x\to a^{+}}f(x)=f(a)$ and ${\lim }_{x\to b^{-}}f(x)=f(b)$

2.9 | Types of Discontinuities

(Definition) Discontinuities $f(x)$ has a removable discontinuity at $x=a$ if ${\lim }_{x\to a}f(x)$ exists but does not equal $f(a)$ $f(x)$ has a jump continuity at $x=a$ if ${\lim }_{x\to a^{-}}f(x)$ and ${\lim }_{x\to a^{+}}$ both exist but are not equal $f(x)$ has an infinite discontinuity at $x=a$ if $f(x)$ has a vertical asymptote at $x=a$ $f(x)$ has an oscillatory discontinuity if it oscillates

2.10 | Intermediate Value Theorem

(Theorem) IVT Suppose $f(x)$ is continuous on $[a,b]$ and that $\alpha \in \mathbb{R}$.is between $f(a)$ and $f(b)$. Then there exists $c\in (a,b)$ such that $f(x)=\alpha$

Ex: show that there is a real number $c$ such that $2^c=3c$.

PROOF: Let $f(x)=2^x-3x$. This is a difference of continuous functions, thus it is continuous. $f(1)=2^1-3(1)=2-3=-1<0$ $f(0)=2^0-3(0)=1-0=1>0$ So $f(a)<0<f(0)$ and $f$ is continuous on $[0,1]$. Therefore, $\exists c\in (0,1)$

Ex: Show that $f(x)=x^3+x-3$ has a real root

PROOF: $f(0)<0<f(2)$ $f(1)=1^3+1-3=-1$ $f(1)<0<f(2)$ $a+0=1,b_0=2$ in $[a_0,b_0]$. There is a root $f(a_0)<0<f(b_0)$ $d=\frac{a_0+b_0}{2}=\frac{1+2}{2}=\frac{3}{2}$ $f(d)=f\left(\frac{3}{2}\right)=\frac{13}{8}>0$ $f(d_0)<0<f(d)$ in $[a_0,d]$ by IVT

$a_1=a_0=1$ $b_1=d=\frac{3}{2}$ root in $[a_1,b_1]$ $d=\frac{a_1+b_1}{2}=\frac{1+\frac{3}{2}}{2}=\frac{5}{4}$ $f(d)=\frac{13}{64}>0$ $f(d_0)<0<f(d)$

$d_2=a_1=1$ $b_2=d=\frac{5}{4}$ root in $[a_2,b_2]$ $d=\frac{a_2+b_2}{2}=\frac{9}{8}$ $f(d)=-\frac{231}{512}>=$ $f(d)<0<f(b_2)$

$a_3=d=\frac{9}{8}$ $b_3=b_2=\frac{5}{4}$ $d=\frac{a_3+b_3}{2}=\frac{19}{16}$ $a_4=1.875$ $b_4=1.25$

etc.

3 | Derivatives

3.1 | Average and Instantaneous Velocity

3.2 | Definition of the Derivative

Definition of the Derivative:

$$\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$

Continuity: A function $f$ is continuous at $a$ if

$$li{m}_{x\to a}f(x)=f(a)$$

But continuity does not imply differentiability - counterexample: $f(x)=|x|$

Symmetric Difference Quotient:

$$f^{\prime }(a)\approx \frac{f(a+h)-f(a-h)}{(a+h)-(a-h)}$$ $$f^{\prime }(a)=\frac{f(a+h)-f(a-h)}{2h}$$

3.3 | Derivatives of Common Functions

L’Hôpital’s Rule

$$li{m}_{x\to a}\frac{f(x)}{g(x)}=li{m}_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

3.4 | Derivative Rules

Power Rule:

$$f^{\prime }(x)=n\cdot x^{n-1}$$

Constant Rule: Let $f(x)=c$

$$f^{\prime }(x)=0$$

Constant Multiple Rule: Let $f(x)=c\cdot g(x)$

$$f^{\prime }(x)=c\cdot g^{\prime }(x)$$

Product / Sum Rule: Let $f(x)=g(x)\pm h(x)$

$$f^{\prime }(x)=g^{\prime }(x)\pm h^{\prime }(x)$$

Power Rule:

$$f^{\prime }(x)=n\cdot x^{n-1}$$

Definition of $e$

$$e=li{m}_{n\to \infty }(1+\frac{1}{n}{)}^n$$

Product Rule: Let $f(x)=g(x)\cdot h(x)$

$$f^{\prime }(x)=g^{\prime }(x)\cdot h(x)+g(x)\cdot h^{\prime }(x)$$

Quotient Rule: Let $F(x)=\frac{f(x)}{g(x)}$

$$F^{\prime }(x)=\frac{f^{\prime }(x)\cdot g(x)-f(x)\cdot g^{\prime }(x)}{(g(x){)}^2}$$

Logs and Exponents

1. $\frac{d}{dx}{e}^x=e^x$
2. $\frac{d}{dx}ln(x)=\frac{1}{x}$
3. $\frac{d}{dx}{b}^x=b^x\cdot ln(b)$
4. $\frac{d}{dx}{\log }_b(x)=\frac{1}{x\cdot ln(b)}$

Trig Functions

1. $\frac{d}{dx}\sin (x)=\cos (x)$
2. $\frac{d}{dx}\cos (x)=-\sin (x)$
3. $\frac{d}{dx}\tan (x)={\sec }^2(x)$
4. $\frac{d}{dx}\cot (x)=-{\csc }^2(x)$
5. $\frac{d}{dx}\sec (x)=\sec (x)\cdot \tan (x)$
6. $\frac{d}{dx}\csc (x)=-\csc (x)\cdot \cot (x)$

Inverse Trig Functions

$$\frac{dy}{dx}arcsin(x)=\frac{1}{\sqrt{1-x^2}}$$ $$\frac{dy}{dx}arccos(x)=-\frac{1}{\sqrt{1-x^2}}$$ $$\frac{dy}{dx}arctan(x)=\frac{1}{1+x^2}$$ $$\frac{dy}{dx}arccot(x)=-\frac{1}{1+x^2}$$ $$\frac{dy}{dx}arccsc(x)=\frac{1}{cs{c}^2(x)}$$

3.5 | Linear Approximation

Local Linear Approximation

$$f(a+h)\approx f(a)+f^{\prime }(a)\ast h$$

3.6 | Newton’s Method

Newton’s Method

$$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$$

3.7 | Derivatives of Inverse Functions

Inverse Function Theorem: For $f^{-1}(x)=g(x)$

$$g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$$

Example: $f(x)=x^2.g(x)=\sqrt{x}$

$$g^{\prime }(x)=\frac{1}{2(x^{\frac{1}{2}})}=\frac{1}{2}{x}^{-1/2}$$

3.8 | Implicit and Logarithmic Differentiation

Implicit Differentiation: Differentiate $y$ with respect to $x$

• Explicitly Defined: $y=f(x)$
• Implicitly Defined: $y=x$

Logarithmic Differentiation: $ln()$ everything

4 | Applications of the Derivative

4.1 | Related Rates

Just relate the rates

4.2 | Extrema

Theorem (Extreme Value Theorem) Let $f(x)$ be continuous on $[a,b]$. Then, $f(x)$ has a global maximum and a global minimum on $[a,b]$.

4.3 | MVT

Theorem (Rolle’s Theorem) If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $x_1=x_2$ for some $a\le x_1\le x_2\le b$, then there exists some point $c\in (a,b)$ such that

$$f^{\prime }(x)=0$$

Theorem (MVT) If f is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists a $c\in (a,b)$ such that

$$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$

Theorem (Bounded Derivative Theorem) If $m\le f^{\prime }(x)\le M$, then

$$f(x)+m(x-a)\le f(x)\le f(a)+M(x-a)$$

Ex: Prove that $\sqrt{51}\in [7+\frac{1}{8},7+\frac{1}{7}]$

$$f(x)=\sqrt{x},a=49,b=64$$ $$f^{\prime }(x)=\frac{1}{2\sqrt{x}}$$

$f^{\prime }(x)$ is decreasing, so

$$m=\frac{1}{2\sqrt{b}}=\frac{1}{16},M=\frac{1}{2\sqrt{a}}=\frac{1}{17}$$

$51\in (49,64)$

$$f^{\prime }(49)+\frac{1}{16}(51-49)\le f(51)\le f(49)+\frac{1}{14}(51-49)$$ $$7+\frac{1}{8}\le \sqrt{51}\le 7+\frac{1}{7}$$

4.4 | Antiderivatives

$$\int a{x}^n=\frac{a}{n+1}{x}^{n+1}$$

4.5 | First Derivatives and Direction

Derivative Tests

• First derivative = increase / decrease
• Second derivative = concave up / down
• Critical points = when $\frac{dy}{dx}=0$
• Inflection points = when $\frac{d^{2}y}{d{x}^2}=0$

4.6 | Second Derivatives and Concavity

See above

4.7 | Classifying Critical Points

First Derivative Test (FDT) and Second Derivative Test (SDT)

4.8 | L’Hopital’s Rule

Theorem (L’Hôpital’s Rule)

$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

4.9 | Curve Sketching

Order of steps

1. Domain and range
2. x and y intercepts
3. Limits at infinity and asymptotes
4. Intervals of increase and decrease
5. Intervals of concave up / down

4.10 | Optimization

Optimization

• Set the desired $\frac{dy}{dx}=0$ and find critical points