User talk:Hayson1991/mathpage

$x=\tan \left(y\right)$

$1=\sec ^{2}\left(y\right)*{\frac {dy}{dx}}$ (Chain rule, derivative of tan=sec^2)

${\frac {1}{\sec ^{2}\left(y\right)}}={\frac {dy}{dx}}$

$\cos ^{2}\left(y\right)={\frac {dy}{dx}}$

${\frac {dy}{dx}}=\cos ^{2}\left(y\right)$

9~

$x^{2}y+xy^{2}=6\,$

$\left(2x*y+x^{2}*{\frac {dy}{dx}}\right)+\left(1*y^{2}+x*2y{\frac {dy}{dx}}\right)=0$

$2xy+x^{2}{\frac {dy}{dx}}+y^{2}+2xy{\frac {dy}{dx}}=0$

$x^{2}{\frac {dy}{dx}}+2xy{\frac {dy}{dx}}=-2xy-y^{2}$

${\frac {dy}{dx}}={\frac {-2xy-y^{2}}{x^{2}+2xy}}$

${\frac {dy}{dx}}=-{\frac {2xy+y^{2}}{x^{2}+2xy}}$

Multiple u's

To Find dy/dx for
$y=2\cos \left(\left(5x\right)^{2}\right)$

The way she explains it

you'll make 3 u's
${\text{Let }}u=2\cos \left(u\right)$

${\text{Let }}u=u^{2}\,$

${\text{Let }}u=5x\,$

Gaaah, help~~

Find ${\frac {dy}{dx}}\,$ then find ${\frac {d^{2}y}{dx^{2}}}\,$

$x^{2}+y^{2}=1\,$

$2x+2y{\frac {dy}{dx}}=0\,$

Find first derivative

${\frac {dy}{dx}}={\frac {-2x}{2y}}\,$

${\frac {dy}{dx}}=-{\frac {x}{y}}\,$

Find second derivative

$2+\left(2{\frac {dy}{dx}}*{\frac {dy}{dx}}+2y*{\frac {d^{2}y}{dx^{2}}}\right)=0\,$

$2\left({\frac {dy}{dx}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,$

$2\left(-{\frac {x}{y}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,$

$2{\frac {x^{2}}{y^{2}}}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,$

$2y{\frac {d^{2}y}{dx^{2}}}=-2-2{\frac {x^{2}}{y^{2}}}\,$

${\frac {d^{2}y}{dx^{2}}}={\frac {-2-2{\frac {x^{2}}{y^{2}}}}{2y}}\,$

${\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y}}-{\frac {x^{2}}{y^{3}}}\,$

${\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}}{y^{3}}}-{\frac {x^{2}}{y^{3}}}\,$

${\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}+x^{2}}{y^{3}}}\,$

${\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y^{3}}}\,$

Clock Problem ~

minute hand

$x=5\cos \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{60}}\right)$

$y=5\sin \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{60}}\right)$

hour hand

$x=4\cos \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{12}}\right)$

$y=4\sin \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{12}}\right)$

Piston speed ~

$x_{w}=r\cos \left(\theta \right)$

$y_{w}=r\sin \left(\theta \right)$

$D_{p}={\sqrt {L^{2}-x_{w}^{2}}}+y_{w}$

${\frac {dD_{p}}{dt}}={\frac {1}{2}}\left(L^{2}-x_{w}^{2}\right)^{-{\frac {1}{2}}}\cdot \left(-2x_{w}{\frac {dx_{w}}{dt}}\right)+{\frac {dy_{w}}{dt}}$

${\frac {dx_{w}}{dt}}=-r\sin \left(\theta \right)\cdot \omega$

${\frac {dy_{w}}{dt}}=r\cos \left(\theta \right)\cdot \omega$

$\theta =t\cdot \omega$

Feon's Question 1~

Solution 1

$y=\tan \left(\arcsin \left(x\right)\right)$

${\frac {dy}{dx}}={\frac {\sec ^{2}\left(\arcsin \left(x\right)\right)}{\sqrt {1-x^{2}}}}$

Solution 2

$y=\tan \left(\arcsin \left(x\right)\right)$

$y={\frac {\sin \left(\arcsin \left(x\right)\right)}{\cos \left(\arcsin \left(x\right)\right)}}$

$y={\frac {x}{\cos \left(\arcsin \left(x\right)\right)}}$

$y=x\sec \left(\arcsin \left(x\right)\right)$

${\frac {dy}{dx}}=\sec \left(\arcsin \left(x\right)\right)+x\cdot {\frac {\sec \left(\arcsin \left(x\right)\right)\tan \left(\arcsin \left(x\right)\right)}{\sqrt {1-x^{2}}}}$

${\frac {dy}{dx}}=\sec \left(\arcsin \left(x\right)\right)+x\cdot {\frac {\sin \left(\arcsin \left(x\right)\right)}{\cos ^{2}\left(\arcsin \left(x\right)\right){\sqrt {1-x^{2}}}}}$

${\frac {dy}{dx}}=\sec \left(\arcsin \left(x\right)\right)+{\frac {x^{2}}{\cos ^{2}\left(\arcsin \left(x\right)\right){\sqrt {1-x^{2}}}}}$

${\frac {dy}{dx}}=\sec \left(\arcsin \left(x\right)\right)+{\frac {x^{2}\sec ^{2}\left(\arcsin \left(x\right)\right)}{\sqrt {1-x^{2}}}}$

Feon's Question 2

$y=x\cdot \operatorname {arcsec} \left(x\right)$

${\frac {dy}{dx}}=\operatorname {arcsec} \left(x\right)+{\frac {x}{\left|x\right|{\sqrt {x^{2}-1}}}}$

${\frac {dy}{dx}}={\begin{cases}\operatorname {arcsec} \left(x\right)+{\frac {1}{\sqrt {x^{2}-1}}}&{\mbox{if }}x>0\\\operatorname {arcsec} \left(x\right)-{\frac {1}{\sqrt {x^{2}-1}}}&{\mbox{if }}x<0\end{cases}}$

Feon's Question 3~~

$y=x\left(\arcsin \left(x\right)\right)^{2}-2x+2{\sqrt {1-x^{2}}}\arcsin \left(x\right)$

$y=x\left(\arcsin \left(x\right)\right)^{2}-2x+2\left(1-x^{2}\right)^{\frac {1}{2}}\arcsin \left(x\right)$

${\frac {dy}{dx}}=\left(\arcsin \left(x\right)\right)^{2}+x\cdot {\frac {2\left(\arcsin \left(x\right)\right)}{\sqrt {1-x^{2}}}}-2+2\cdot {\frac {1}{2}}\left(1-x^{2}\right)^{-{\frac {1}{2}}}\cdot -2x\cdot \arcsin \left(x\right)+{\frac {2{\sqrt {1-x^{2}}}}{\sqrt {1-x^{2}}}}$

${\frac {dy}{dx}}=\left(\arcsin \left(x\right)\right)^{2}+x\cdot {\frac {2\left(\arcsin \left(x\right)\right)}{\sqrt {1-x^{2}}}}-2+2\cdot {\frac {1}{2{\sqrt {1-x^{2}}}}}\cdot -2x\cdot \arcsin \left(x\right)+{\frac {2{\sqrt {1-x^{2}}}}{\sqrt {1-x^{2}}}}$

${\frac {dy}{dx}}=\arcsin ^{2}\left(x\right)+{\frac {2x\arcsin \left(x\right)}{\sqrt {1-x^{2}}}}-{\frac {2x\arcsin \left(x\right)}{\sqrt {1-x^{2}}}}-2+2$

${\frac {dy}{dx}}=\arcsin ^{2}\left(x\right)$

Last Part

$u\cdot v=2{\sqrt {1-x^{2}}}\cdot \arcsin \left(x\right)$

$u=2{\sqrt {1-x^{2}}}$

$v=\arcsin \left(x\right)$

$u'=2\cdot {\frac {1}{2{\sqrt {1-x^{2}}}}}\cdot -2x$

$u'={\frac {-2x}{\sqrt {1-x^{2}}}}$

$v'={\frac {1}{\sqrt {1-x^{2}}}}$

$u'v+v'u={\frac {-2x}{\sqrt {1-x^{2}}}}\cdot \arcsin \left(x\right)+{\frac {1}{\sqrt {1-x^{2}}}}\cdot 2{\sqrt {1-x^{2}}}$

$u'v+v'u=-{\frac {2x\arcsin \left(x\right)}{\sqrt {1-x^{2}}}}+2$

ln derivative

$y=\ln {\sqrt {\frac {x-1}{x+1}}}$
${\frac {dy}{dx}}={\frac {1}{\sqrt {\frac {x-1}{x+1}}}}\cdot {\frac {1}{2{\sqrt {\frac {x-1}{x+1}}}}}\cdot {\frac {1\left(x+1\right)-1\left(x-1\right)}{\left(x+1\right)^{2}}}$
${\frac {dy}{dx}}={\sqrt {\frac {x+1}{x-1}}}\cdot \left({\frac {1}{2}}\cdot {\sqrt {\frac {x+1}{x-1}}}\right)\cdot {\frac {2}{\left(x+1\right)^{2}}}$
${\frac {dy}{dx}}={\frac {x+1}{2x-2}}\cdot {\frac {2}{\left(x+1\right)^{2}}}$
${\frac {dy}{dx}}={\frac {1}{\left(x-1\right)\left(x+1\right)}}$
${\frac {dy}{dx}}={\frac {1}{x^{2}-1}}$

ln derivative 2~

$y=\ln {\left(x+{\sqrt {4+x^{2}}}\right)}$
${\frac {dy}{dx}}={\frac {1}{\left(x+{\sqrt {4+x^{2}}}\right)}}\cdot \left(1+{\frac {1}{2{\sqrt {4+x^{2}}}}}\cdot 2x\right)$

Difference

For the original function f:

1

${\frac {1}{b-a}}\int _{a}^{b}{fdx}$

2

${\frac {f(b)-f(a)}{b-a}}$

TingTing's Physics Problem ~

$\,x\,$	$\,y\,$
$x={\mbox{given}}\,$	$y={\mbox{given}}\,$
$t=\,$	$t=\,$
$v=v_{0}\cos \left(\theta \right)\,$	$v=v_{0}\sin \left(\theta \right)\,$
$a=0\,$	$a=-9.81\,$

$y=v_{0}\sin \left(\theta \right)t+{\frac {1}{2}}at^{2}\,$

$x=v_{0}\cos \left(\theta \right)t\,$
${\frac {x}{v_{0}\cos \left(\theta \right)}}=t\,$
$t={\frac {x}{v_{0}\cos \left(\theta \right)}}\,$

$y=v_{0}\sin \left(\theta \right)t+{\frac {1}{2}}a\left({\frac {x}{v_{0}\cos \left(\theta \right)}}\right)^{2}\,$

How to do logs without a calculator.

Memorization

There are 2 constants you have to memorize:

${\text{1) }}{\frac {1}{2}}\approx \log \left(3.16\right)$

${\text{2) }}{\begin{cases}\log \left(e\right)\approx 0.434\\\,\,\,\,\,{\text{ or }}\\{\frac {1}{\log \left(e\right)}}\approx 2.304\end{cases}}$

Method

Example 1

$\log \left(1234.34599\right)$

$=\log \left(1.23434599\times 10^{3}\right)$

$=\log \left(1.23434599\right)+\log \left(10^{3}\right)$

$=\log \left(1.23434599\right)+3$

${\text{Since }}1.23<3.16{\text{, }}\log \left(1234.34599\right){\text{ is between }}3.0{\text{ and }}3.5$

${\text{And }}\log \left(1234.34599\right)\approx 3+1.23\div 3.16\times {\frac {1}{2}}\approx 3.195$

Example 2

$\log \left(53.32423\right)$

$=\log \left(5.332423\times 10^{1}\right)$

$=\log \left(5.332423\right)+\log \left(10^{1}\right)$

$=\log \left(5.332423\right)+1$

${\text{Since }}5.33>3.16{\text{, }}\log \left(53.32423\right){\text{ is between }}1.5{\text{ and }}2.0$

${\text{And }}\log \left(53.32423\right)\approx 1+5.33\div 3.16\times {\frac {1}{2}}\approx 1.843$

Example 3

$\log \left(0.00003942\right)$

$=\log \left(3.942\times 10^{-5}\right)$

$=\log \left(3.942\right)+\log \left(10^{-5}\right)$

$=\log \left(3.942\right)-5$

${\text{Since }}3.942>3.16{\text{, }}\log \left(0.00003942\right){\text{ is between }}-4.0{\text{ and }}-4.5$

${\text{And }}\log \left(0.00003942\right)\approx -5+3.942\div 3.16\times {\frac {1}{2}}\approx -4.376$

Example 4

$\ln \left(234.213\right)$

$=\log _{e}\left(234.213\right)$

$={\frac {\log \left(234.213\right)}{\log \left(e\right)}}$

$={\frac {1}{\log \left(e\right)}}\times \log \left(2.34213\times 10^{2}\right)$

$={\frac {1}{\log \left(e\right)}}\times \log \left(2.34213\right)+{\frac {2}{\log \left(e\right)}}$

$={\frac {1}{0.434}}\times \log \left(2.34213\right)+{\frac {2}{0.434}}$

$=2.304\times \log \left(2.34213\right)+2\times 2.304$

$=2.304\times \log \left(2.34213\right)+4.608$

$\approx 2.304\times \left(2.34213\div 3.16\times {\frac {1}{2}}\right)+4.608$

$\approx 5.857$

TingTing's Problem

Because of $y'''=-8y+5\,$ and $y=e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\,$:

$y'''=-8\left(e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\right)+5\,$

$y=e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\,$

$y'=-2e^{-2x}+3Bx^{2}+2Cx+D\,$

$y''=4e^{-2x}+3\cdot 2Bx+2C\,$

$y'''=-8e^{-2x}+3\cdot 2B=-8e^{-2x}+6B\,$

Since $y'''=-8e^{-2x}+6B\,$ and $y'''=-8\left(e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\right)+5\,$:

$-8e^{-2x}+6B=-8\left(e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\right)+5\,$

$6B=-8Bx^{3}-8Cx^{2}-8Dx-8E+5\,$

$0x^{3}+0x^{2}+0x+6B=-8Bx^{3}-8Cx^{2}-8Dx-8E+5\,$

Therefore:
${\begin{cases}0=-8B\,\\0=-8C\,\\0=-8D\,\\6B=-8E+5\,\end{cases}}$

${\begin{cases}B=0\,\\C=0\,\\D=0\,\\E={\frac {5}{8}}\,\end{cases}}$

$y=e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\,$

$y=e^{-2x}+{\frac {5}{8}}\,$