## Multivariable Calculus Equations

$\lim\limits_{x \to c} f(x) = L$

$\lim \limits_{x \to c} b = b$
${\mathop {\lim }\limits_{x \to c} x = c}$
${\mathop {\lim }\limits_{x \to c} kf(x) = k\cdot\mathop {\lim }\limits_{x \to c} f(x) = kL}$
${\mathop {\lim }\limits_{x \to c} [f(x) + g(x)] = \mathop {\lim }\limits_{x \to c} f(x) + \mathop {\lim }\limits_{x \to c} g(x) = L + M}$
${\mathop {\lim }\limits_{x \to c} [f(x) - g(x)] = \mathop {\lim }\limits_{x \to c} f(x) - \mathop {\lim }\limits_{x \to c} g(x) = L - M}$
${\mathop {\lim }\limits_{x \to c} [f(x)g(x)] = \mathop {\lim }\limits_{x \to c} f(x)\mathop {\lim }\limits_{x \to c} g(x) = LM}$
$\mathop {\lim }\limits_{x \to c} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to c} f(x)}}{{\mathop {\lim }\limits_{x \to c} g(x)}} = \frac{L}{M}$ provided $M \ne 0$

$\mathop {\lim }\limits_{x \to c} f{(x)^n} = {\left( {\mathop {\lim }\limits_{x \to c} f(x)} \right)^n} = {L^n} \hfill$
$g(x) \leqslant f(x) \leqslant h(x) \hfill$
$\mathop {\lim }\limits_{x \to c} g(x) = \mathop {\lim }\limits_{x \to c} h(x) = L \hfill$
$\mathop {\lim }\limits_{x \to c} f(x) = L \hfill$

$\mathop {\lim }\limits_{n \to \infty } ({a_n} + {b_n}) = \mathop {\lim }\limits_{n \to \infty } {a_n} + \mathop {\lim }\limits_{n \to \infty } {b_n} \hfill$
$\mathop {\lim }\limits_{n \to \infty } ({a_n} - {b_n}) = \mathop {\lim }\limits_{n \to \infty } {a_n} - \mathop {\lim }\limits_{n \to \infty } {b_n} \hfill$
$\mathop {\lim }\limits_{n \to \infty } c{a_n} = c\mathop {\lim }\limits_{n \to \infty } {a_n} \hfill$
$\mathop {\lim }\limits_{n \to \infty } c = c \hfill$
$\mathop {\lim }\limits_{n \to \infty } ({a_n}{b_n}) = \mathop {\lim }\limits_{n \to \infty } {a_n}\cdot\mathop {\lim }\limits_{n \to \infty } {b_n} \hfill$
$\mathop {\lim }\limits_{x \to \infty } \frac{{{a_n}}}{{{b_n}}} = \frac{{\mathop {\lim }\limits_{n \to \infty } {a_n}}}{{\mathop {\lim }\limits_{n \to \infty } {b_n}}} \hfill$ if $\mathop {\lim }\limits_{n \to \infty } {b_n} \ne 0$
$\mathop {\lim }\limits_{n \to \infty } a_n^p = {[\mathop {\lim }\limits_{n \to \infty } {a_n}]^p} \hfill$ if $p > 0$ and ${a_n}>0$

If $\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0 \hfill$, then $\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$

If $\mathop {\lim }\limits_{n \to \infty } {a_n} = L$ and the function $f$ is continuous at $L$, then $\mathop {\lim }\limits_{n \to \infty } f({a_n}) = f(L)$

$\mathop {\lim }\limits_{n \to \infty } {r^n} = \left\{ {\begin{array}{*{20}{l}} 0&{ - 1 < r < 1} \\ 1&{r = 1} \end{array}} \right.$

The geometric series $\sum _{n=1}^{\infty } ar^{n-1}=a+ar+ar^2+\cdots$ is convergent if $\left| r \right| < 1$ and its sum is $\sum _{n=1}^{\infty } ar^{n-1}=\frac{a}{1-r}$

If $\left| r \right| \leqslant \ 1$, the geometric series is divergent.

If the series $\sum _{n=1}^{\infty }{a_n}$ is convergent, then $\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$.

If $\mathop {\lim }\limits_{n \to \infty } {a_n}$ does not exist or if $\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0$, then the series $\sum _{n=1}^{\infty } {a_n}$ is divergent.

If $\sum{a_n}$ and $\sum {{b_n}}$ are convergent series, then so are the series $\sum {c{a_n}}$ (where $c$ is a constant), $\sum {({a_n} + {b_n})}$, and $\sum {({a_n} - {b_n})}$, and

$\sum \limits_{n=1}^{\infty} ca_n=c\sum\limits_{n=1}^{\infty } a_n$
$\sum _{n=1}^{\infty } (a_n+b_n)=\sum _{n=1}^{\infty } a_n+\sum _{n=1}^{\infty } b_n$
$\sum _{n=1}^{\infty } (a_n-b_n)=\sum _{n=1}^{\infty } a_n-\sum _{n=1}^{\infty } b_n$

## Math Equations 2

$\int \sec xdx=\ln |\sec x+\tan x|+C=\ln \left|\tan \left(\frac{1}{2}x+\frac{1}{4}\pi \right)\right|+C$
$\int \csc xdx=-\ln |\csc x+\cot x|+C=\ln \left|\tan \left(\frac{1}{2}x\right)\right|+C$
$\int \cot xdx=\ln |\sin x|+C$
$\int \sec ^2kxdx=\frac{1}{k}\tan kx+C$
$\int \csc ^2kxdx=-\frac{1}{k}\cot kx+C$
$\int \cot ^2kxdx=-x-\frac{1}{k}\cot kx+C$
$\int \sec x\tan xdx=\sec x+C$
$\int \sec x\csc xdx=\ln |\tan x|+C$

## Math Equations

Calculus

$\int cf(x)dx=c\int f(x)dx$
$\int f(x)+g(x)dx=\int f(x)dx+\int g(x)dx$
$\int f(x)-g(x)dx=\int f(x)dx-\int g(x)dx$
$\int udv\, =uv-\int vdu$

$\int dx=x+C$
$\int adx=ax+C$
$\int x^ndx=\frac{1}{n+1}x^{n+1}+C$ if $n\neq -1$
$\int \frac{1}{x}dx=\ln |x|+C$
$\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+C$ if $a\neq 0$

$\int \sin xdx=-\cos x+C$
$\int \cos xdx=\sin x+C$
$\int \tan xdx=\ln |\sec x|+C$
$\int \sin ^2xdx=\frac{1}{2}x-\frac{1}{4}\sin 2x+C$
$\int \cos ^2xdx=\frac{1}{2}x+\frac{1}{4}\sin 2x+C$
$\int \tan ^2xdx=\tan (x)-x+C$
$\int \sin ^nxdx=-\frac{\sin ^{n-1}x\cos x}{n}+\frac{n-1}{n}$
$\int \sin ^{n-2}xdx+C$ for $n>0$
$\int \cos ^nxdx=-\frac{\cos ^{n-1}x\sin x}{n}+\frac{n-1}{n}$
$\int \cos ^{n-2}xdx+C$ for $n>0$
$\int \tan ^nxdx=\frac{1}{(n-1)}\tan ^{n-1}x-\int \tan ^{n-2}xdx+C$ for $n\neq 1)$

$\int \sec xdx=\ln |\sec x+\tan x|+C=\ln \left|\tan \left(\frac{1}{2}x+\frac{1}{4}\pi \right)\right|+C$
$\int \csc xdx=-\ln |\csc x+\cot x|+C=\ln \left|\tan \left(\frac{1}{2}x\right)\right|+C$
$\int \cot xdx=\ln |\sin x|+C$
$\int \sec ^2kxdx=\frac{1}{k}\tan kx+C$
$\int \csc ^2kxdx=-\frac{1}{k}\cot kx+C$
$\int \cot ^2kxdx=-x-\frac{1}{k}\cot kx+C$
$\int \sec x\tan xdx=\sec x+C$
$\int \sec x\csc xdx=\ln |\tan x|+C$

$\int \sec ^nxdx=\frac{\sec ^{n-1}x\sin x}{n-1}+\frac{n-2}{n-1}$
$\int \sec ^{n-2}xdx+C$ for $n\neq 1$
$\int \csc ^nxdx=-\frac{\csc ^{n-1}x\cos x}{n-1}+\frac{n-2}{n-1}$
$\int \csc ^{n-2}xdx+C$ for $n\neq 1$
$\int \cot ^nxdx=-\frac{1}{n-1}\cot ^{n-1}x-\int \cot ^{n-2}xdx+C$ for $n\neq 1$
$\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin (x)+C$
$\int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin (x/a)+C$ if $a\neq 0$
$\int \frac{1}{1+x^2}dx=\arctan (x)+C$
$\int \frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan (x/a)+C$ if $a\neq 0$
$\int e^xdx=e^x+C$
$\int e^{ax}dx=\frac{1}{a}e^{ax}+C$ if $a\neq 0$
$\int a^xdx=\frac{1}{\ln a}a^x+C$ if $a>0,a\neq 1$
$\int \ln xdx=x\ln x-x+C$

$\int \arcsin (x)dx=x\arcsin (x)+\sqrt{1-x^2}+C$
$\int \arccos (x)dx=x\arccos (x)-\sqrt{1-x^2}+C$
$\int \arctan (x)dx=x\arctan (x)-\frac{1}{2}\ln (1+x^2)+C$

$\frac{d}{dx}(f+g)=\frac{df}{dx}+\frac{dg}{dx}$
$\frac{d}{dx}(cf)=c\frac{df}{dx}$
$\frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}$
$\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\frac{df}{dx}-f\frac{dg}{dx}}{g^2}$

$\frac{d}{dx}(c)=0$
$\frac{d}{dx}x=1$
$\frac{d}{dx}x^n=nx^{n-1}$
$\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$
$\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$
$\frac{d}{dx}(c_nx^n+c_{n-1}x^{n-1}+c_{n-2}x^{n-2}+\cdots +c_2x^2+c_1x+c_0)=nc_nx^{n-1}+(n-1)c_{n-1}x^{n-2}+(n-2)c_{n-2}x^{n-3}+\cdots +2c_2x+c_1$

$\frac{d}{dx}\sin (x)=\cos (x)$
$\frac{d}{dx}\cos (x)=-\sin (x)$
$\frac{d}{dx}\tan (x)=\sec ^2(x)$
$\frac{d}{dx}\cot (x)=-\csc ^2(x)$
$\frac{d}{dx}\sec (x)=\sec (x)\tan (x)\frac{d}{dx}\csc (x)=-\csc (x)\cot (x)$

$\frac{d}{dx}e^x=e^x\frac{d}{dx}a^x=a^x\ln (a)$ if $a>0$
$\frac{d}{dx}\ln (x)=\frac{1}{x}\frac{d}{dx}\log _a(x)=\frac{1}{x\ln (a)}$ if $a>0,a\neq 1$
${({f^g})^\prime } = {\left( {{e^{g\ln f}}} \right)^\prime } = {f^g}\left( {f'\frac{g}{f} + g'\ln f} \right),\qquad f > 0 \hfill$
${({c^f})^\prime } = {\left( {{e^{f\ln c}}} \right)^\prime } = f'{c^f}\ln c \hfill$

$\frac{d}{dx}\text{arcsinx}=\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\text{arccosx}=-\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\text{arctanx}=\frac{1}{1+x^2}$
$\frac{d}{dx}\text{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}$
$\frac{d}{dx}\text{arccot}x=\frac{-1}{1+x^2}$
$\frac{d}{dx}\text{arccsc}x=\frac{-1}{|x|\sqrt{x^2-1}}$

$\frac{d}{dx}\sinh x=\cosh x$
$\frac{d}{dx}\cosh x=\sinh x$
$\frac{d}{dx}\tanh x=\text{sech}^2x$
$\frac{d}{dx}\text{sech}x=-\tanh x\text{sech}x$
$\frac{d}{dx}\coth x=-\text{csch}^2x$
$\frac{d}{dx}\text{csch}x=-\coth x\text{csch}x$
$\frac{d}{dx}\sinh ^{-1}x=\frac{1}{\sqrt{x^2+1}}$
$\frac{d}{dx}\cosh ^{-1}x=\frac{-1}{\sqrt{x^2-1}}$
$\frac{d}{dx}\tanh ^{-1}x=\frac{1}{1-x^2}$
$\frac{d}{dx}\text{sech}^{-1}x=\frac{1}{x\sqrt{1-x^2}}$
$\frac{d}{dx}\coth ^{-1}x=\frac{-1}{1-x^2}$
$\frac{d}{dx}\text{csch}^{-1}x=\frac{-1}{|x|\sqrt{1+x^2}}$

$\sum _{i=N}^M f(i)=f(N)+f(N+1)+f(N+2)+\cdots +f(M)$

$\sum _{i=1}^n c=c+c+$$+c=nc,c\in \mathbb{R}$
$\sum _{i=1}^n i=1+2+3+$$+n=\frac{n(n+1)}{2}$
$\sum _{i=1}^n i^2=1^2+2^2+3^2+$$+n^2=\frac{n(n+1)(2n+1)}{6}$
$\sum _{i=1}^n i^3=1^3+2^3+3^3+$$+n^3=\frac{n^2(n+1)^2}{4}$

$\tan (x)=\frac{\sin x}{\cos x}$
$\sec (x)=\frac{1}{\cos x}$
$\cot (x)=\frac{\cos x}{\sin x}=\frac{1}{\tan x}$
$\csc (x)=\frac{1}{\sin x}$

$\sin ^2x+\cos ^2x=1\,$
$1 + {\tan ^2}(x) = {\sec ^2}x\; \hfill$
$1 + {\cot ^2}(x) = {\csc ^2}x\; \hfill$

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

$\sin (x+y)=\sin x\cos y+\cos x\sin y$
$\sin (x-y)=\sin x\cos y-\cos x\sin y$
$\cos (x+y)=\cos x\cos y-\sin x\sin y$
$\cos (x-y)=\cos x\cos y+\sin x\sin y$
$\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$
$\sin x-\sin y=2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$
$\cos x+\cos y=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$
$\cos x-\cos y=-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$
$\tan x+\tan y=\frac{\sin (x+y)}{\cos x\cos y}$
$\tan x-\tan y=\frac{\sin (x-y)}{\cos x\cos y}$
$\cot x+\cot y=\frac{\sin (x+y)}{\sin x\sin y}$
$\cot x-\cot y=\frac{-\sin (x-y)}{\sin x\sin y}$

$\cos (x)\cos (y)=\frac{\cos (x+y)+\cos (x-y)}{2}$
$\sin (x)\sin (y)=\frac{\cos (x-y)-\cos (x+y)}{2}\,$
$\sin (x)\cos (y)=\frac{\sin (x+y)+\sin (x-y)}{2}\,$
$\cos (x)\sin (y)=\frac{\sin (x+y)-\sin (x-y)}{2}\,$

## Mathematica

I have been working with Mathematica for about a month now and I think that it is an excellent program to use and is essential for anyone who is serious about math after high school.Â  I have been working on making lessons specific to regular math classes.Â  I am hoping that these lessons can create a new style of learning math that will utilize the technology that we have to a greater extent and will allow teachers to go through some of the material faster.Â  For example, while studying different methods of factoring I began to realize that none of these methods can find irrational numbers and become more difficult to use as the polynomial gets bigger.Â  Mathematica does not have this limitation.Â  Because of this, I feel that it would be best to remove this part from the curriculum and focus more on what the roots of a polynomial mean.

The downloads are broken up into two parts.Â  The first part contains notebook files which you need mathematica for.Â  The second part has cdf files like the ones contained in the demonstrations on Mathematica’s website.Â  For this you need a cdf player which you can download for free on their website.

Notebook Files (need rar)

Critical Points (view pdf)

Algebra (view pdf)

Distance Between Skew Lines (view pdf)

Minimum, Lower Quartile, Median, Upper Quartile, Maximum

## Free Educational Resources

Julie Harland’s Math Videos (Megaupload links, 8 parts, 3.61 GB)

Chycho’s Math Series (magnet links)

The Language of Mathematics Series I

The Language of Mathematics Series II

The Language of Mathematics Series IIIa

The Language of Mathematics Series IIIb

Project Gutenberg

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## Arkansas Standardized Tests

Here are a collection of End of Course Examinations from Arkansas Department of Education.