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<math>x = \tan\left(y\right)</math><br /><br />
<math>1 = \sec^2\left(y\right)*\frac{dy}{dx}</math> (Chain rule, derivative of tan=sec^2)<br /><br />
<math>\frac{1}{\sec^2\left(y\right)} = \frac{dy}{dx}</math><br /><br />
<math>\cos^2\left(y\right) = \frac{dy}{dx}</math><br /><br />
<math>\frac{dy}{dx} = \cos^2\left(y\right)</math><br /><br />

== 9~ ==
<math>x^{2}y + xy^2 = 6\,</math><br /><br />
<math>\left(2x*y + x^{2}*\frac{dy}{dx}\right) + \left(1*y^2 + x*2y\frac{dy}{dx}\right) = 0</math><br /><br />
<math>2xy + x^{2}\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0</math><br /><br />
<math>x^{2}\frac{dy}{dx} + 2xy\frac{dy}{dx} = -2xy - y^2</math><br /><br />
<math>\frac{dy}{dx} = \frac{-2xy - y^2}{x^{2} + 2xy}</math><br /><br />
<math>\frac{dy}{dx} = -\frac{2xy + y^2}{x^{2} + 2xy}</math><br /><br />

== Multiple u's ==

To Find dy/dx for<br />
<math>y = 2\cos\left(\left(5x\right)^2\right)</math><br /><br />
===The way she explains it===
you'll make 3 u's<br />
<math>\text{Let }u = 2\cos\left(u\right)</math><br /><br />
<math>\text{Let }u = u^2\,</math><br /><br />
<math>\text{Let }u = 5x\,</math><br /><br />

== Gaaah, help~~ ==


Find <math>\frac{dy}{dx}\,</math> then find <math>\frac{d^2y}{dx^2}\,</math> <br /><br />

<math>x^2 + y^2 = 1\,</math><br /><br />
<math>2x + 2y\frac{dy}{dx} = 0\,</math><br /><br />
===Find first derivative===
<math>\frac{dy}{dx} = \frac{-2x}{2y}\,</math><br /><br />
<math>\frac{dy}{dx} = -\frac{x}{y}\,</math><br /><br />
===Find second derivative===
<math>2 + \left(2\frac{dy}{dx}*\frac{dy}{dx} + 2y*\frac{d^{2}y}{dx^2}\right) = 0\,</math><br /><br />
<math>2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br />
<math>2\left(-\frac{x}{y}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br />
<math>2\frac{x^2}{y^2} + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br />
<math>2y\frac{d^{2}y}{dx^2} = -2-2\frac{x^2}{y^2}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = \frac{-2-2\frac{x^2}{y^2}}{2y}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{1}{y} - \frac{x^2}{y^3}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{y^2}{y^3} - \frac{x^2}{y^3}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{y^2+x^2}{y^3}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{1}{y^3}\,</math><br /><br />

== Clock Problem ~ ==

===minute hand===
<math>x=5\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{60}\right)</math><br /><br />
<math>y=5\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{60}\right)</math><br /><br />

===hour hand===
<math>x=4\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br />
<math>y=4\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br />

== Piston speed ~ ==

<math>x_w = r\cos\left(\theta\right)</math><br /><br />
<math>y_w = r\sin\left(\theta\right)</math><br /><br />
<math>D_p = \sqrt{L^2-x_w^2}+y_w</math><br /><br />
<math>\frac{dD_p}{dt} = \frac{1}{2}\left(L^2-x_w^2\right)^{-\frac{1}{2}}\cdot\left(-2x_w\frac{dx_w}{dt}\right)+\frac{dy_w}{dt}</math><br /><br />
<math>\frac{dx_w}{dt} = -r\sin\left(\theta\right)\cdot\omega</math><br /><br />
<math>\frac{dy_w}{dt} = r\cos\left(\theta\right)\cdot\omega</math><br /><br />
<math>\theta = t\cdot\omega</math><br /><br />

== Feon's Question 1~ ==
=== Solution 1 ===
<math>y = \tan\left(\arcsin\left(x\right)\right)</math><br /><br />
<math>\frac{dy}{dx} = \frac{\sec^2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}</math><br /><br />
=== Solution 2 ===
<math>y = \tan\left(\arcsin\left(x\right)\right)</math><br /><br />
<math>y = \frac{\sin\left(\arcsin\left(x\right)\right)}{\cos\left(\arcsin\left(x\right)\right)}</math><br /><br />
<math>y = \frac{x}{\cos\left(\arcsin\left(x\right)\right)}</math><br /><br />
<math>y = x\sec\left(\arcsin\left(x\right)\right)</math><br /><br />
<math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + x\cdot\frac{\sec\left(\arcsin\left(x\right)\right)\tan\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}</math><br /><br />
<math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + x\cdot\frac{\sin\left(\arcsin\left(x\right)\right)}{\cos^2\left(\arcsin\left(x\right)\right)\sqrt{1-x^2}}</math><br /><br />
<math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + \frac{x^2}{\cos^2\left(\arcsin\left(x\right)\right)\sqrt{1-x^2}}</math><br /><br />
<math>\frac{dy}{dx} = \sec\left(\arcsin\left(x\right)\right) + \frac{x^2\sec^2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}</math><br /><br />

==Feon's Question 2==
<math>y = x\cdot\arcsec\left(x\right)</math><br /><br />
<math>\frac{dy}{dx} = \arcsec\left(x\right) + \frac{x}{\left|x\right|\sqrt{x^2-1}}</math><br /><br />
<math>\frac{dy}{dx} = \begin{cases}
\arcsec\left(x\right) + \frac{1}{\sqrt{x^2-1}}& \mbox{if }x>0 \\
\arcsec\left(x\right) - \frac{1}{\sqrt{x^2-1}}& \mbox{if }x<0
\end{cases}</math><br /><br />
==Feon's Question 3~~==
<math>y = x\left(\arcsin\left(x\right)\right)^2-2x+2\sqrt{1-x^2}\arcsin\left(x\right)</math><br /><br />
<math>y = x\left(\arcsin\left(x\right)\right)^2-2x+2\left(1-x^2\right)^\frac{1}{2}\arcsin\left(x\right)</math><br /><br />
<math>\frac{dy}{dx} = \left(\arcsin\left(x\right)\right)^2+x \cdot \frac{2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}-2+2\cdot \frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}} \cdot -2x \cdot \arcsin\left(x\right) + \frac{2\sqrt{1-x^2}}{\sqrt{1-x^2}}</math><br /><br /><math>\frac{dy}{dx} = \left(\arcsin\left(x\right)\right)^2+x \cdot \frac{2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}-2+2\cdot \frac{1}{2\sqrt{1-x^2}} \cdot -2x \cdot \arcsin\left(x\right) + \frac{2\sqrt{1-x^2}}{\sqrt{1-x^2}}</math><br /><br />
<math>\frac{dy}{dx} = \arcsin^2\left(x\right)+\frac{2x\arcsin\left(x\right)}{\sqrt{1-x^2}}-\frac{2x\arcsin\left(x\right)}{\sqrt{1-x^2}}-2+2</math><br /><br />
<math>\frac{dy}{dx} = \arcsin^2\left(x\right)</math><br /><br />
===Last Part===
<math>u \cdot v=2\sqrt{1-x^2}\cdot\arcsin\left(x\right)</math><br /><br />
<math>u = 2\sqrt{1-x^2}</math><br /><br />
<math>v = \arcsin\left(x\right)</math><br /><br />
<math>u' = 2\cdot\frac{1}{2\sqrt{1-x^2}}\cdot-2x</math><br /><br />
<math>u' = \frac{-2x}{\sqrt{1-x^2}}</math><br /><br />
<math>v' = \frac{1}{\sqrt{1-x^2}}</math><br /><br />
<math>u'v + v'u = \frac{-2x}{\sqrt{1-x^2}} \cdot \arcsin\left(x\right) + \frac{1}{\sqrt{1-x^2}} \cdot 2\sqrt{1-x^2}</math><br /><br />
<math>u'v + v'u = -\frac{2x\arcsin\left(x\right)}{\sqrt{1-x^2}} + 2</math><br /><br />

== ln derivative ==

<math>y = \ln{\sqrt{\frac{x-1}{x+1}}}</math><br />
<math>\frac{dy}{dx} = \frac{1}{\sqrt{\frac{x-1}{x+1}}}\cdot\frac{1}{2\sqrt{\frac{x-1}{x+1}}}\cdot\frac{1\left(x+1\right)-1\left(x-1\right)}{\left(x+1\right)^2}</math><br />
<math>\frac{dy}{dx} = \sqrt{\frac{x+1}{x-1}}\cdot\left(\frac{1}{2}\cdot\sqrt{\frac{x+1}{x-1}}\right)\cdot\frac{2}{\left(x+1\right)^2}</math><br />
<math>\frac{dy}{dx} = \frac{x+1}{2x-2}\cdot\frac{2}{\left(x+1\right)^2}</math><br />
<math>\frac{dy}{dx} = \frac{1}{\left(x-1\right)\left(x+1\right)}</math><br />
<math>\frac{dy}{dx} = \frac{1}{x^2-1}</math><br />

== ln derivative 2~ ==

<math>y = \ln{\left(x+\sqrt{4+x^2}\right)}</math><br />
<math>\frac{dy}{dx} = \frac{1}{\left(x+\sqrt{4+x^2}\right)}\cdot\left(1+\frac{1}{2\sqrt{4+x^2}}\cdot 2x\right)</math><br />

== Difference ==

For the original function f:
===1===
<math>\frac{1}{b-a}\int_a^b{fdx}</math>
===2===
<math>\frac{f(b)-f(a)}{b-a}</math>

== TingTing's Physics Problem ~ ==

{| class="wikitable" border="1"
|-
! <math>\,x\,</math>
! <math>\,y\,</math>
|-
| <math>x = \mbox{given} \,</math>
| <math>y = \mbox{given} \,</math>
|-
| <math>t = \,</math>
| <math>t = \,</math>
|-
| <math>v = v_0 \cos\left(\theta\right) \,</math>
| <math>v = v_0 \sin\left(\theta\right) \,</math>
|-
| <math>a = 0 \,</math>
| <math>a = -9.81 \,</math>
|}

<math>y = v_0 \sin\left(\theta\right) t + \frac{1}{2} a t^2\,</math><br /><br />
<math>x = v_0 \cos\left(\theta\right) t\,</math><br />
<math>\frac{x}{v_0 \cos\left(\theta\right)} = t\,</math><br />
<math>t = \frac{x}{v_0 \cos\left(\theta\right)}\,</math><br /><br />
<math>y = v_0 \sin\left(\theta\right) t + \frac{1}{2} a \left(\frac{x}{v_0 \cos\left(\theta\right)}\right)^2\,</math><br />

== How to do logs without a calculator. ==

===Memorization===
There are 2 constants you have to memorize:<br /><br />
<math>\text{1) }\frac{1}{2}\approx\log\left(3.16\right)</math><br /><br />
<math>\text{2) }\begin{cases}
\log\left(e\right)\approx 0.434 \\
\,\,\,\,\,\text{ or } \\
\frac{1}{\log\left(e\right)}\approx 2.304
\end{cases}</math>

===Method===
====Example 1====
<math>\log\left(1234.34599\right)</math><br /><br />
<math>=\log\left(1.23434599\times 10^3\right)</math><br /><br />
<math>=\log\left(1.23434599\right)+\log\left(10^3\right)</math><br /><br />
<math>=\log\left(1.23434599\right)+3</math><br /><br />
:<math>\text{Since }1.23 < 3.16\text{, }\log\left(1234.34599\right)\text{ is between }3.0\text{ and }3.5</math><br /><br />
:<math>\text{And }\log\left(1234.34599\right)\approx 3 + 1.23 \div 3.16 \times \frac{1}{2}\approx 3.195</math><br /><br />
====Example 2====
<math>\log\left(53.32423\right)</math><br /><br />
<math>=\log\left(5.332423\times 10^1\right)</math><br /><br />
<math>=\log\left(5.332423\right)+\log\left(10^1\right)</math><br /><br />
<math>=\log\left(5.332423\right)+1</math><br /><br />
:<math>\text{Since }5.33 > 3.16\text{, }\log\left(53.32423\right)\text{ is between }1.5\text{ and }2.0</math><br /><br />
:<math>\text{And }\log\left(53.32423\right)\approx 1 + 5.33 \div 3.16 \times \frac{1}{2}\approx 1.843</math><br /><br />
====Example 3====
<math>\log\left(0.00003942\right)</math><br /><br />
<math>=\log\left(3.942\times 10^{-5}\right)</math><br /><br />
<math>=\log\left(3.942\right)+\log\left(10^{-5}\right)</math><br /><br />
<math>=\log\left(3.942\right)-5</math><br /><br />
:<math>\text{Since }3.942 > 3.16\text{, }\log\left(0.00003942\right)\text{ is between }-4.0\text{ and }-4.5</math><br /><br />
:<math>\text{And }\log\left(0.00003942\right)\approx -5 + 3.942 \div 3.16 \times \frac{1}{2}\approx -4.376</math><br /><br />
====Example 4====
<math>\ln\left(234.213\right)</math><br /><br />
<math>=\log_e\left(234.213\right)</math><br /><br />
<math>=\frac{\log\left(234.213\right)}{\log\left(e\right)}</math><br /><br />
<math>=\frac{1}{\log\left(e\right)}\times\log\left(2.34213\times 10^2\right)</math><br /><br />
<math>=\frac{1}{\log\left(e\right)}\times\log\left(2.34213\right)+\frac{2}{\log\left(e\right)}</math><br /><br />
<math>=\frac{1}{0.434}\times\log\left(2.34213\right)+\frac{2}{0.434}</math><br /><br />
<math>=2.304\times\log\left(2.34213\right)+2\times 2.304</math><br /><br />
<math>=2.304\times\log\left(2.34213\right)+4.608</math><br /><br />
<math>\approx 2.304\times\left(2.34213\div 3.16\times\frac{1}{2}\right)+4.608</math><br /><br />
<math>\approx 5.857</math>

== TingTing's Problem ==

Because of <math>y''' = -8 y + 5\,</math> and <math>y = e^{-2x} + B x^3 + C x^2 + D x + E\,</math>:<br /><br />
<math>y''' = -8 \left( e^{-2x} + B x^3 + C x^2 + D x + E\right) + 5\,</math><br /><br /><br />

<math>y = e^{-2x} + B x^3 + C x^2 + D x + E\,</math><br /><br />
<math>y' = -2 e^{-2x} + 3B x^2 + 2C x + D\,</math><br /><br />
<math>y'' = 4 e^{-2x} + 3 \cdot 2B x + 2C\,</math><br /><br />
<math>y''' = -8 e^{-2x} + 3 \cdot 2B = -8 e^{-2x} + 6B\,</math><br /><br /><br />

Since <math>y''' = -8 e^{-2x} + 6B\,</math> and <math>y''' = -8 \left( e^{-2x} + B x^3 + C x^2 + D x + E\right) + 5\,</math>:<br /><br />
<math>-8 e^{-2x} + 6B = -8 \left( e^{-2x} + B x^3 + C x^2 + D x + E\right) + 5\,</math><br /><br />
<math>6B = -8 B x^3 -8 C x^2 -8 D x -8E + 5\,</math><br /><br />
<math>0 x^3 + 0 x^2 + 0 x + 6B = -8 B x^3 -8 C x^2 -8 D x -8E + 5\,</math><br /><br /><br />

Therefore:<br />
<math>
\begin{cases}
0 = -8 B\, \\
0 = -8 C\, \\
0 = -8 D\, \\
6B = -8E + 5\,
\end{cases}
</math><br /><br />
<math>
\begin{cases}
B = 0\, \\
C = 0\, \\
D = 0\, \\
E = \frac{5}{8}\,
\end{cases}
</math><br /><br /><br />

<math>y = e^{-2x} + B x^3 + C x^2 + D x + E\,</math><br /><br />
<math>y = e^{-2x} + \frac{5}{8}\,</math><br /><br />

== Kelly's problem ==

<math>\int^0_k \left( 2kx-5k \right) dx = k^2</math><br /><br />
<math>\left^{x=0}_{x=k} = k^2</math><br /><br />
<math>\left( 0-0 \right)-\left( k\cdot k^2-5k\cdot k \right) = k^2</math><br /><br />
<math>-\left( k\cdot k^2-5k\cdot k \right) = k^2</math><br /><br />
<math>k^3-5k^2= -k^2 \,</math><br /><br />
<math>k^3= 4k^2 \,</math><br /><br />
<math>k=4 \,</math><br /><br />


== Taylor Polynomial problem ==
Use the Taylor polynomial of minimal degree to approximate <math>ln(1.4)</math> to accuracy .01.

Step 1: Make remainder smaller than .01 How?

Latest revision as of 11:03, 6 July 2012