< User talk:Hayson1991 Browse history interactively ← Previous edit Next edit → Content deleted Content addedVisual Wikitext Inline Revision as of 21:40, 30 December 2008
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}}}
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}}\,}
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)}
(Chain rule, derivative of tan=sec^2)
then find 
