< User talk:Hayson1991 Browse history interactively ← Previous edit Next edit → Content deleted Content addedVisual Wikitext Inline Revision as of 02:40, 28 January 2010
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)}
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}\,}
How to do logs without a calculator.
Memorization
There are 2 constants you have to memorize:
1)
1
2
≈
log
(
3.16
)
{\displaystyle {\text{1) }}{\frac {1}{2}}\approx \log \left(3.16\right)}
2)
{
log
(
e
)
≈
0.434
or
1
log
(
e
)
≈
2.304
{\displaystyle {\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
(
1234.34599
)
{\displaystyle \log \left(1234.34599\right)}
=
log
(
1.23434599
×
10
3
)
{\displaystyle =\log \left(1.23434599\times 10^{3}\right)}
=
log
(
1.23434599
)
+
log
(
10
3
)
{\displaystyle =\log \left(1.23434599\right)+\log \left(10^{3}\right)}
=
log
(
1.23434599
)
+
3
{\displaystyle =\log \left(1.23434599\right)+3}
Since
1.23
<
3.16
,
log
(
1234.34599
)
is between
3.0
and
3.5
{\displaystyle {\text{Since }}1.23<3.16{\text{, }}\log \left(1234.34599\right){\text{ is between }}3.0{\text{ and }}3.5}
And
log
(
1234.34599
)
≈
3
+
1.23
÷
3.16
×
1
2
≈
3.195
{\displaystyle {\text{And }}\log \left(1234.34599\right)\approx 3+1.23\div 3.16\times {\frac {1}{2}}\approx 3.195}
Example 2
log
(
53.32423
)
{\displaystyle \log \left(53.32423\right)}
=
log
(
5.332423
×
10
1
)
{\displaystyle =\log \left(5.332423\times 10^{1}\right)}
=
log
(
5.332423
)
+
log
(
10
1
)
{\displaystyle =\log \left(5.332423\right)+\log \left(10^{1}\right)}
=
log
(
5.332423
)
+
1
{\displaystyle =\log \left(5.332423\right)+1}
Since
5.33
>
3.16
,
log
(
53.32423
)
is between
1.5
and
2.0
{\displaystyle {\text{Since }}5.33>3.16{\text{, }}\log \left(53.32423\right){\text{ is between }}1.5{\text{ and }}2.0}
And
log
(
53.32423
)
≈
1
+
5.33
÷
3.16
×
1
2
≈
1.843
{\displaystyle {\text{And }}\log \left(53.32423\right)\approx 1+5.33\div 3.16\times {\frac {1}{2}}\approx 1.843}
Example 3
log
(
0.00003942
)
{\displaystyle \log \left(0.00003942\right)}
=
log
(
3.942
×
10
−
5
)
{\displaystyle =\log \left(3.942\times 10^{-5}\right)}
=
log
(
3.942
)
+
log
(
10
−
5
)
{\displaystyle =\log \left(3.942\right)+\log \left(10^{-5}\right)}
=
log
(
3.942
)
−
5
{\displaystyle =\log \left(3.942\right)-5}
Since
3.942
>
3.16
,
log
(
0.00003942
)
is between
−
4.0
and
−
4.5
{\displaystyle {\text{Since }}3.942>3.16{\text{, }}\log \left(0.00003942\right){\text{ is between }}-4.0{\text{ and }}-4.5}
And
log
(
0.00003942
)
≈
−
5
+
3.942
÷
3.16
×
1
2
≈
−
4.376
{\displaystyle {\text{And }}\log \left(0.00003942\right)\approx -5+3.942\div 3.16\times {\frac {1}{2}}\approx -4.376}
Example 4
ln
(
234.213
)
{\displaystyle \ln \left(234.213\right)}
=
log
e
(
234.213
)
{\displaystyle =\log _{e}\left(234.213\right)}
=
log
(
234.213
)
log
(
e
)
{\displaystyle ={\frac {\log \left(234.213\right)}{\log \left(e\right)}}}
=
1
log
(
e
)
×
log
(
2.34213
×
10
2
)
{\displaystyle ={\frac {1}{\log \left(e\right)}}\times \log \left(2.34213\times 10^{2}\right)}
=
1
log
(
e
)
×
log
(
2.34213
)
+
2
log
(
e
)
{\displaystyle ={\frac {1}{\log \left(e\right)}}\times \log \left(2.34213\right)+{\frac {2}{\log \left(e\right)}}}
=
1
0.434
×
log
(
2.34213
)
+
2
0.434
{\displaystyle ={\frac {1}{0.434}}\times \log \left(2.34213\right)+{\frac {2}{0.434}}}
=
2.304
×
log
(
2.34213
)
+
2
×
2.304
{\displaystyle =2.304\times \log \left(2.34213\right)+2\times 2.304}
=
2.304
×
log
(
2.34213
)
+
4.608
{\displaystyle =2.304\times \log \left(2.34213\right)+4.608}
≈
2.304
×
(
2.34213
÷
3.16
×
1
2
)
+
4.608
{\displaystyle \approx 2.304\times \left(2.34213\div 3.16\times {\frac {1}{2}}\right)+4.608}
≈
5.857
{\displaystyle \approx 5.857}
TingTing's Problem
Because of
y
‴
=
−
8
y
+
5
{\displaystyle y'''=-8y+5\,}
y
=
e
−
2
x
+
B
x
3
+
C
x
2
+
D
x
+
E
{\displaystyle y=e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\,}
y
‴
=
−
8
(
e
−
2
x
+
B
x
3
+
C
x
2
+
D
x
+
E
)
+
5
{\displaystyle y'''=-8\left(e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\right)+5\,}
y
=
e
−
2
x
+
B
x
3
+
C
x
2
+
D
x
+
E
{\displaystyle y=e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\,}
y
′
=
−
2
e
−
2
x
+
3
B
x
2
+
2
C
x
+
D
{\displaystyle y'=-2e^{-2x}+3Bx^{2}+2Cx+D\,}
y
″
=
4
e
−
2
x
+
3
⋅
2
B
x
+
2
C
{\displaystyle y''=4e^{-2x}+3\cdot 2Bx+2C\,}
y
‴
=
−
8
e
−
2
x
+
3
⋅
2
B
=
−
8
e
−
2
x
+
6
B
{\displaystyle y'''=-8e^{-2x}+3\cdot 2B=-8e^{-2x}+6B\,}
Since
y
‴
=
−
8
e
−
2
x
+
6
B
{\displaystyle y'''=-8e^{-2x}+6B\,}
y
‴
=
−
8
(
e
−
2
x
+
B
x
3
+
C
x
2
+
D
x
+
E
)
+
5
{\displaystyle y'''=-8\left(e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\right)+5\,}
−
8
e
−
2
x
+
6
B
=
−
8
(
e
−
2
x
+
B
x
3
+
C
x
2
+
D
x
+
E
)
+
5
{\displaystyle -8e^{-2x}+6B=-8\left(e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\right)+5\,}
6
B
=
−
8
B
x
3
−
8
C
x
2
−
8
D
x
−
8
E
+
5
{\displaystyle 6B=-8Bx^{3}-8Cx^{2}-8Dx-8E+5\,}
0
x
3
+
0
x
2
+
0
x
+
6
B
=
−
8
B
x
3
−
8
C
x
2
−
8
D
x
−
8
E
+
5
{\displaystyle 0x^{3}+0x^{2}+0x+6B=-8Bx^{3}-8Cx^{2}-8Dx-8E+5\,}
Therefore:
{
0
=
−
8
B
0
=
−
8
C
0
=
−
8
D
6
B
=
−
8
E
+
5
{\displaystyle {\begin{cases}0=-8B\,\\0=-8C\,\\0=-8D\,\\6B=-8E+5\,\end{cases}}}
{
B
=
0
C
=
0
D
=
0
E
=
5
8
{\displaystyle {\begin{cases}B=0\,\\C=0\,\\D=0\,\\E={\frac {5}{8}}\,\end{cases}}}
y
=
e
−
2
x
+
B
x
3
+
C
x
2
+
D
x
+
E
{\displaystyle y=e^{-2x}+Bx^{3}+Cx^{2}+Dx+E\,}
y
=
e
−
2
x
+
5
8
{\displaystyle y=e^{-2x}+{\frac {5}{8}}\,}
Kelly's problem
∫
k
0
(
2
k
x
−
5
k
)
d
x
=
k
2
{\displaystyle \int _{k}^{0}\left(2kx-5k\right)dx=k^{2}}
[
k
x
2
−
5
k
x
]
x
=
k
x
=
0
=
k
2
{\displaystyle \left_{x=k}^{x=0}=k^{2}}
(
0
−
0
)
−
(
k
⋅
k
2
−
5
k
⋅
k
)
=
k
2
{\displaystyle \left(0-0\right)-\left(k\cdot k^{2}-5k\cdot k\right)=k^{2}}
−
(
k
⋅
k
2
−
5
k
⋅
k
)
=
k
2
{\displaystyle -\left(k\cdot k^{2}-5k\cdot k\right)=k^{2}}
k
3
−
5
k
2
=
−
k
2
{\displaystyle k^{3}-5k^{2}=-k^{2}\,}
k
3
=
4
k
2
{\displaystyle k^{3}=4k^{2}\,}
k
=
4
{\displaystyle k=4\,}
Taylor Polynomial problem
Use the Taylor polynomial of minimal degree to approximate
l
n
(
1.4
)
{\displaystyle ln(1.4)}
Step 1: Make remainder smaller than .01 How? —Preceding unsigned comment added by 128.61.76.165 (talk ) 02:39, 28 January 2010 (UTC)
(Chain rule, derivative of tan=sec^2)
then find 
and 
:
and 
to accuracy .01.