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This is an old revision of this page, as edited by 24.40.202.246 (talk) at 22:15, 1 March 2009 (→TingTing's Physics Problem). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

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x = tan ⁡ ( y ) {\displaystyle x=\tan \left(y\right)}

1 = sec 2 ⁡ ( y ) ∗ d y d x {\displaystyle 1=\sec ^{2}\left(y\right)*{\frac {dy}{dx}}} (Chain rule, derivative of tan=sec^2)

1 sec 2 ⁡ ( y ) = d y d x {\displaystyle {\frac {1}{\sec ^{2}\left(y\right)}}={\frac {dy}{dx}}}

cos 2 ⁡ ( y ) = d y d x {\displaystyle \cos ^{2}\left(y\right)={\frac {dy}{dx}}}

d y d x = cos 2 ⁡ ( y ) {\displaystyle {\frac {dy}{dx}}=\cos ^{2}\left(y\right)}

9~

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

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

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

x 2 d y d x + 2 x y d y d x = − 2 x y − y 2 {\displaystyle x^{2}{\frac {dy}{dx}}+2xy{\frac {dy}{dx}}=-2xy-y^{2}}

d y d x = − 2 x y − y 2 x 2 + 2 x y {\displaystyle {\frac {dy}{dx}}={\frac {-2xy-y^{2}}{x^{2}+2xy}}}

d y d x = − 2 x y + y 2 x 2 + 2 x y {\displaystyle {\frac {dy}{dx}}=-{\frac {2xy+y^{2}}{x^{2}+2xy}}}

Multiple u's

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

The way she explains it

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

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

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

Gaaah, help~~

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

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

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

Find first derivative

d y d x = − 2 x 2 y {\displaystyle {\frac {dy}{dx}}={\frac {-2x}{2y}}\,}

d y d x = − x y {\displaystyle {\frac {dy}{dx}}=-{\frac {x}{y}}\,}

Find second derivative

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

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

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

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

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

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

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

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

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

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

Clock Problem ~

minute hand

x = 5 cos ⁡ ( π 2 − t ⋅ 2 π 60 ) {\displaystyle x=5\cos \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{60}}\right)}

y = 5 sin ⁡ ( π 2 − t ⋅ 2 π 60 ) {\displaystyle y=5\sin \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{60}}\right)}

hour hand

x = 4 cos ⁡ ( π 2 − t ⋅ 2 π 12 ) {\displaystyle x=4\cos \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{12}}\right)}

y = 4 sin ⁡ ( π 2 − t ⋅ 2 π 12 ) {\displaystyle y=4\sin \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{12}}\right)}

Piston speed ~

x w = r cos ⁡ ( θ ) {\displaystyle x_{w}=r\cos \left(\theta \right)}

y w = r sin ⁡ ( θ ) {\displaystyle y_{w}=r\sin \left(\theta \right)}

D p = L 2 − x w 2 + y w {\displaystyle D_{p}={\sqrt {L^{2}-x_{w}^{2}}}+y_{w}}

d D p d t = 1 2 ( L 2 − x w 2 ) − 1 2 ⋅ ( − 2 x w d x w d t ) + d y w d t {\displaystyle {\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}}}

d x w d t = − r sin ⁡ ( θ ) ⋅ ω {\displaystyle {\frac {dx_{w}}{dt}}=-r\sin \left(\theta \right)\cdot \omega }

d y w d t = r cos ⁡ ( θ ) ⋅ ω {\displaystyle {\frac {dy_{w}}{dt}}=r\cos \left(\theta \right)\cdot \omega }

θ = t ⋅ ω {\displaystyle \theta =t\cdot \omega }

Feon's Question 1~

Solution 1

y = tan ⁡ ( arcsin ⁡ ( x ) ) {\displaystyle y=\tan \left(\arcsin \left(x\right)\right)}

d y d x = sec 2 ⁡ ( arcsin ⁡ ( x ) ) 1 − x 2 {\displaystyle {\frac {dy}{dx}}={\frac {\sec ^{2}\left(\arcsin \left(x\right)\right)}{\sqrt {1-x^{2}}}}}

Solution 2

y = tan ⁡ ( arcsin ⁡ ( x ) ) {\displaystyle y=\tan \left(\arcsin \left(x\right)\right)}

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

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

y = x sec ⁡ ( arcsin ⁡ ( x ) ) {\displaystyle y=x\sec \left(\arcsin \left(x\right)\right)}

d y d x = sec ⁡ ( arcsin ⁡ ( x ) ) + x ⋅ sec ⁡ ( arcsin ⁡ ( x ) ) tan ⁡ ( arcsin ⁡ ( x ) ) 1 − x 2 {\displaystyle {\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}}}}}

d y d x = sec ⁡ ( arcsin ⁡ ( x ) ) + x ⋅ sin ⁡ ( arcsin ⁡ ( x ) ) cos 2 ⁡ ( arcsin ⁡ ( x ) ) 1 − x 2 {\displaystyle {\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}}}}}}

d y d x = sec ⁡ ( arcsin ⁡ ( x ) ) + x 2 cos 2 ⁡ ( arcsin ⁡ ( x ) ) 1 − x 2 {\displaystyle {\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}}}}}}

d y d x = sec ⁡ ( arcsin ⁡ ( x ) ) + x 2 sec 2 ⁡ ( arcsin ⁡ ( x ) ) 1 − x 2 {\displaystyle {\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 ⋅ arcsec ⁡ ( x ) {\displaystyle y=x\cdot \operatorname {arcsec} \left(x\right)}

d y d x = arcsec ⁡ ( x ) + x | x | x 2 − 1 {\displaystyle {\frac {dy}{dx}}=\operatorname {arcsec} \left(x\right)+{\frac {x}{\left|x\right|{\sqrt {x^{2}-1}}}}}

d y d x = { arcsec ⁡ ( x ) + 1 x 2 − 1 if  x > 0 arcsec ⁡ ( x ) − 1 x 2 − 1 if  x < 0 {\displaystyle {\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 ( arcsin ⁡ ( x ) ) 2 − 2 x + 2 1 − x 2 arcsin ⁡ ( x ) {\displaystyle y=x\left(\arcsin \left(x\right)\right)^{2}-2x+2{\sqrt {1-x^{2}}}\arcsin \left(x\right)}

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

d y d x = ( arcsin ⁡ ( x ) ) 2 + x ⋅ 2 ( arcsin ⁡ ( x ) ) 1 − x 2 − 2 + 2 ⋅ 1 2 ( 1 − x 2 ) − 1 2 ⋅ − 2 x ⋅ arcsin ⁡ ( x ) + 2 1 − x 2 1 − x 2 {\displaystyle {\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}}}}}

d y d x = ( arcsin ⁡ ( x ) ) 2 + x ⋅ 2 ( arcsin ⁡ ( x ) ) 1 − x 2 − 2 + 2 ⋅ 1 2 1 − x 2 ⋅ − 2 x ⋅ arcsin ⁡ ( x ) + 2 1 − x 2 1 − x 2 {\displaystyle {\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}}}}}

d y d x = arcsin 2 ⁡ ( x ) + 2 x arcsin ⁡ ( x ) 1 − x 2 − 2 x arcsin ⁡ ( x ) 1 − x 2 − 2 + 2 {\displaystyle {\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}

d y d x = arcsin 2 ⁡ ( x ) {\displaystyle {\frac {dy}{dx}}=\arcsin ^{2}\left(x\right)}

Last Part

u ⋅ v = 2 1 − x 2 ⋅ arcsin ⁡ ( x ) {\displaystyle u\cdot v=2{\sqrt {1-x^{2}}}\cdot \arcsin \left(x\right)}

u = 2 1 − x 2 {\displaystyle u=2{\sqrt {1-x^{2}}}}

v = arcsin ⁡ ( x ) {\displaystyle v=\arcsin \left(x\right)}

u ′ = 2 ⋅ 1 2 1 − x 2 ⋅ − 2 x {\displaystyle u'=2\cdot {\frac {1}{2{\sqrt {1-x^{2}}}}}\cdot -2x}

u ′ = − 2 x 1 − x 2 {\displaystyle u'={\frac {-2x}{\sqrt {1-x^{2}}}}}

v ′ = 1 1 − x 2 {\displaystyle v'={\frac {1}{\sqrt {1-x^{2}}}}}

u ′ v + v ′ u = − 2 x 1 − x 2 ⋅ arcsin ⁡ ( x ) + 1 1 − x 2 ⋅ 2 1 − x 2 {\displaystyle 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 = − 2 x arcsin ⁡ ( x ) 1 − x 2 + 2 {\displaystyle u'v+v'u=-{\frac {2x\arcsin \left(x\right)}{\sqrt {1-x^{2}}}}+2}

ln derivative

y = ln ⁡ x − 1 x + 1 {\displaystyle y=\ln {\sqrt {\frac {x-1}{x+1}}}}
d y d x = 1 x − 1 x + 1 ⋅ 1 2 x − 1 x + 1 ⋅ 1 ( x + 1 ) − 1 ( x − 1 ) ( x + 1 ) 2 {\displaystyle {\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}}}}
d y d x = x + 1 x − 1 ⋅ ( 1 2 ⋅ x + 1 x − 1 ) ⋅ 2 ( x + 1 ) 2 {\displaystyle {\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}}}}
d y d x = x + 1 2 x − 2 ⋅ 2 ( x + 1 ) 2 {\displaystyle {\frac {dy}{dx}}={\frac {x+1}{2x-2}}\cdot {\frac {2}{\left(x+1\right)^{2}}}}
d y d x = 1 ( x − 1 ) ( x + 1 ) {\displaystyle {\frac {dy}{dx}}={\frac {1}{\left(x-1\right)\left(x+1\right)}}}
d y d x = 1 x 2 − 1 {\displaystyle {\frac {dy}{dx}}={\frac {1}{x^{2}-1}}}

ln derivative 2~

y = ln ⁡ ( x + 4 + x 2 ) {\displaystyle y=\ln {\left(x+{\sqrt {4+x^{2}}}\right)}}
d y d x = 1 ( x + 4 + x 2 ) ⋅ ( 1 + 1 2 4 + x 2 ⋅ 2 x ) {\displaystyle {\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

1 b − a ∫ a b f d x {\displaystyle {\frac {1}{b-a}}\int _{a}^{b}{fdx}}

2

f ( b ) − f ( a ) b − a {\displaystyle {\frac {f(b)-f(a)}{b-a}}}

TingTing's Physics Problem ~

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

y = v 0 sin ⁡ ( θ ) t + 1 2 a t 2 {\displaystyle y=v_{0}\sin \left(\theta \right)t+{\frac {1}{2}}at^{2}\,}

x = v 0 cos ⁡ ( θ ) t {\displaystyle x=v_{0}\cos \left(\theta \right)t\,}
x v 0 cos ⁡ ( θ ) = t {\displaystyle {\frac {x}{v_{0}\cos \left(\theta \right)}}=t\,}
t = x v 0 cos ⁡ ( θ ) {\displaystyle t={\frac {x}{v_{0}\cos \left(\theta \right)}}\,}

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