Derivative of a function f at a number a [ edit | edit source ] We denote the derivative of a function
f
{\displaystyle f}
a
{\displaystyle a}
f
′
(
a
)
{\displaystyle f'(a)\,\!}
The derivative of a function
f
{\displaystyle f}
a
{\displaystyle a}
limit (if it exists):
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
{\displaystyle f'(a)=\lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}}
An analagous equation can be defined by letting
x
=
(
a
+
h
)
{\displaystyle x=(a+h)}
h
=
(
x
−
a
)
{\displaystyle h=(x-a)}
x
{\displaystyle x}
a
{\displaystyle a}
h
{\displaystyle h}
0
{\displaystyle 0}
f
′
(
a
)
=
lim
x
→
a
f
(
x
)
−
f
(
a
)
x
−
a
{\displaystyle f'(a)=\lim _{x\rightarrow a}{\frac {f(x)-f(a)}{x-a}}}
As the slope of a tangent line [ edit | edit source ] Given a function
y
=
f
(
x
)
{\displaystyle y=f(x)\,\!}
f
′
(
a
)
{\displaystyle f'(a)\,\!}
f
(
x
)
{\displaystyle f(x)}
x
=
a
{\displaystyle x=a}
Find the equation of the tangent line to
y
=
x
2
{\displaystyle y=x^{2}}
x
=
1
{\displaystyle x=1}
To find the slope of the tangent, we let
y
=
f
(
x
)
{\displaystyle y=f(x)}
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
⇒
lim
h
→
0
(
a
+
h
)
2
−
(
a
)
2
h
⇒
lim
h
→
0
a
2
+
2
a
h
+
h
2
−
a
2
h
⇒
lim
h
→
0
h
(
2
a
+
h
)
h
⇒
lim
h
→
0
(
2
a
+
h
)
{\displaystyle f'(a)=\lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}(a+h)^{2}-(a)^{2}}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}a^{2}+2ah+h^{2}-a^{2}}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}h(2a+h)}{h}}\Rightarrow \lim _{h\rightarrow 0}{\color {Blue}(2a+h)}}
It can be seen that as
h
{\displaystyle h}
0
{\displaystyle 0}
f
′
(
a
)
=
2
a
{\displaystyle f'(a)={\color {Blue}2a}\,\!}
1
{\displaystyle 1}
a
{\displaystyle a}
f
′
(
1
)
=
2
(
1
)
⇒
2
{\displaystyle f'({\color {Red}1})=2({\color {Red}1})\Rightarrow {\color {Red}2}}
So our preliminary equation for the tangent line is
y
=
2
x
+
b
{\displaystyle y={\color {Red}2}x+b}
(
1
,
1
)
{\displaystyle (1,1)}
b
{\displaystyle b}
1
=
2
(
1
)
+
b
⇒
−
1
=
b
{\displaystyle {\color {Red}1}=2({\color {Red}1})+b\Rightarrow -1=b}
So our final equation is
y
=
2
x
−
1
{\displaystyle y=2x-1\,\!}
The derivative of a function
f
(
x
)
{\displaystyle f(x)}
a
{\displaystyle a}
f
(
x
)
{\displaystyle f(x)}
x
=
a
{\displaystyle x=a}
The derivative as a function [ edit | edit source ] So far we have only examined the derivative of a function
f
{\displaystyle f}
a
{\displaystyle a}
a
{\displaystyle a}
x
{\displaystyle x}
f
{\displaystyle f}
x
{\displaystyle x}
f
{\displaystyle f}
a
{\displaystyle a}
number , whereas the derivative of
f
{\displaystyle f}
function .
Likewise to the derivative of
f
{\displaystyle f}
a
{\displaystyle a}
f
(
x
)
{\displaystyle f(x)}
f
′
(
x
)
{\displaystyle f'(x)\,\!}
The derivative of the function
f
{\displaystyle f}
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
{\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}
Also,
f
′
(
x
)
=
lim
h
→
x
f
(
x
)
−
f
(
h
)
x
−
h
{\displaystyle f'(x)=\lim _{h\rightarrow x}{\frac {f(x)-f(h)}{x-h}}}
or
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
−
h
)
2
h
{\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x-h)}{2h}}}
Consider the sequences:
y
=
x
2
;
y
+
Δ
y
=
(
x
+
h
)
2
=
x
2
+
2
x
h
+
h
2
{\displaystyle y=x^{2};\ y+\Delta y=(x+h)^{2}=x^{2}+2xh+h^{2}}
y
=
x
3
;
y
+
Δ
y
=
(
x
+
h
)
3
=
x
3
+
3
x
2
h
+
3
x
h
2
+
h
3
{\displaystyle y=x^{3};\ y+\Delta y=(x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}}
y
=
x
4
;
y
+
Δ
y
=
(
x
+
h
)
4
=
x
4
+
4
x
3
h
+
6
x
2
h
2
+
4
x
h
3
+
h
4
{\displaystyle y=x^{4};\ y+\Delta y=(x+h)^{4}=x^{4}+4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}}
y
=
x
5
;
y
+
Δ
y
=
(
x
+
h
)
5
=
x
5
+
5
x
4
h
+
10
x
3
h
2
+
10
x
2
h
3
+
5
x
h
4
+
h
5
{\displaystyle y=x^{5};\ y+\Delta y=(x+h)^{5}=x^{5}+5x^{4}h+10x^{3}h^{2}+10x^{2}h^{3}+5xh^{4}+h^{5}}
y
=
x
6
;
y
+
Δ
y
=
(
x
+
h
)
6
=
x
6
+
6
x
5
h
+
{\displaystyle y=x^{6};\ y+\Delta y=(x+h)^{6}=x^{6}+6x^{5}h\ +}
h
2
{\displaystyle h^{2}}
y
=
x
n
;
y
+
Δ
y
=
(
x
+
Δ
x
)
n
=
x
n
+
n
x
n
−
1
Δ
x
+
{\displaystyle y=x^{n};\ y+\Delta y=(x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x\ +}
(
Δ
x
)
2
{\displaystyle (\Delta x)^{2}}
Δ
y
=
n
x
n
−
1
Δ
x
+
{\displaystyle \Delta y=nx^{n-1}\Delta x\ +}
(
Δ
x
)
2
{\displaystyle (\Delta x)^{2}}
Therefore:
Δ
y
Δ
x
=
n
x
n
−
1
+
{\displaystyle {\frac {\Delta y}{\Delta x}}=nx^{n-1}\ +}
Δ
x
{\displaystyle \Delta x}
d
y
d
x
=
lim
Δ
x
→
0
Δ
y
Δ
x
{\displaystyle {\frac {dy}{dx}}=\lim _{\Delta x\rightarrow 0}{\frac {\Delta y}{\Delta x}}}
d
(
x
n
)
d
x
=
n
x
n
−
1
{\displaystyle {\frac {d(x^{n})}{dx}}=nx^{n-1}}
x
{\displaystyle x}
x
{\displaystyle x}
n
{\displaystyle n}
Later it will be shown that this is valid for all real
n
{\displaystyle n}
Derivative of cubic function [ edit | edit source ]
Graph of cubic function illustrating use of associated quadratic.
x
==
7.5
,
{\displaystyle x==7.5,}
x
{\displaystyle x}
f
(
x
)
=
f
(
7.5
)
.
{\displaystyle f(x)=f(7.5).}
x
==
3
,
{\displaystyle x==3,}
x
{\displaystyle x}
f
(
x
)
=
f
(
3
)
.
{\displaystyle f(x)=f(3).}
x
==
p
,
{\displaystyle x==p,}
x
{\displaystyle x}
f
(
x
)
=
f
(
p
)
.
{\displaystyle f(x)=f(p).}
(
p
,
f
(
p
)
)
{\displaystyle (p,f(p))}
In the diagram there is a stationary point at
x
=
p
.
{\displaystyle x=p.}
x
=
p
,
{\displaystyle x=p,}
x
{\displaystyle x}
f
(
x
)
=
f
(
p
)
.
{\displaystyle f(x)=f(p).}
Aim of this section is to derive the condition that produces exactly 2 values of
x
.
{\displaystyle x.}
product
of linear function and quadratic.
The associated quadratic when
x
=
p
:
{\displaystyle x=p:}
A
=
a
{\displaystyle A=a}
B
=
A
p
+
b
=
a
p
+
b
{\displaystyle B=Ap+b=ap+b}
C
=
B
p
+
c
=
a
p
p
+
b
p
+
c
.
{\displaystyle C=Bp+c=app+bp+c.}
g
(
x
)
=
a
x
2
+
(
a
p
+
b
)
x
+
(
a
p
2
+
b
p
+
c
)
{\displaystyle g(x)=ax^{2}+(ap+b)x+(ap^{2}+bp+c)}
(
x
−
p
)
.
{\displaystyle (x-p).}
This division gives a quotient of
a
x
+
(
2
a
p
+
b
)
{\displaystyle ax+(2ap+b)}
h
(
p
)
=
3
a
p
2
+
2
b
p
+
c
.
{\displaystyle h(p)=3ap^{2}+2bp+c.}
Factor
(
x
−
p
)
{\displaystyle (x-p)}
g
(
x
)
{\displaystyle g(x)}
h
(
p
)
=
0
,
{\displaystyle h(p)=0,}
When
x
=
p
,
h
(
x
)
=
3
a
x
2
+
2
b
x
+
c
,
{\displaystyle x=p,\ h(x)=3ax^{2}+2bx+c,}
f
(
x
)
.
{\displaystyle f(x).}
Derivative of quartic function [ edit | edit source ] Let
y
=
u
{\displaystyle y=u}
⋅
v
{\displaystyle v}
u
=
U
(
x
)
;
v
=
V
(
x
)
{\displaystyle u=U(x);\ v=V(x)}
y
+
Δ
y
=
(
u
+
Δ
u
)
{\displaystyle y+\Delta y=(u+\Delta u)}
⋅
(
v
+
Δ
v
)
=
u
{\displaystyle (v+\Delta v)=u}
⋅
v
+
u
{\displaystyle v+u}
⋅
Δ
v
+
v
{\displaystyle \Delta v+v}
⋅
Δ
u
+
Δ
u
{\displaystyle \Delta u+\Delta u}
⋅
Δ
v
{\displaystyle \Delta v}
Δ
y
=
u
{\displaystyle \Delta y=u}
⋅
Δ
v
+
v
{\displaystyle \Delta v+v}
⋅
Δ
u
+
Δ
u
{\displaystyle \Delta u+\Delta u}
⋅
Δ
v
{\displaystyle \Delta v}
Δ
y
Δ
x
=
u
{\displaystyle {\frac {\Delta y}{\Delta x}}=u}
⋅
Δ
v
Δ
x
+
v
{\displaystyle {\frac {\Delta v}{\Delta x}}+v}
⋅
Δ
u
Δ
x
+
(
Δ
u
)
(
Δ
v
)
Δ
x
{\displaystyle {\frac {\Delta u}{\Delta x}}+{\frac {(\Delta u)(\Delta v)}{\Delta x}}}
d
y
d
x
=
u
{\displaystyle {\frac {dy}{dx}}=u}
⋅
d
v
d
x
+
v
{\displaystyle {\frac {dv}{dx}}+v}
⋅
d
u
d
x
{\displaystyle {\frac {du}{dx}}}
Let
y
=
x
−
n
=
1
x
n
{\displaystyle y=x^{-n}={\frac {1}{x^{n}}}}
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
y
x
n
=
1
{\displaystyle yx^{n}=1}
Differentiate both sides.
y
n
x
n
−
1
+
x
n
d
y
d
x
=
0
{\displaystyle ynx^{n-1}+x^{n}{\frac {dy}{dx}}=0}
d
y
d
x
=
−
y
n
x
n
−
1
x
n
=
−
n
x
n
−
1
x
n
x
n
=
−
n
x
n
−
1
x
−
2
n
=
−
n
x
−
n
−
1
{\displaystyle {\frac {dy}{dx}}=-{\frac {ynx^{n-1}}{x^{n}}}=-{\frac {nx^{n-1}}{x^{n}x^{n}}}=-nx^{n-1}x^{-2n}=-nx^{-n-1}}
This shows that
d
(
x
n
)
d
x
{\displaystyle {\frac {d(x^{n})}{dx}}}
n
{\displaystyle n}
Used where
y
=
U
(
x
)
V
(
x
)
{\displaystyle y={\frac {U(x)}{V(x)}}}
y
=
u
v
{\displaystyle y={\frac {u}{v}}}
y
v
=
u
{\displaystyle yv=u}
y
v
′
+
v
y
′
=
u
′
{\displaystyle yv'+vy'=u'}
v
y
′
=
u
′
−
y
v
′
=
u
′
−
u
v
′
v
{\displaystyle vy'=u'-yv'=u'-{\frac {uv'}{v}}}
y
′
=
u
′
v
−
u
v
′
v
2
=
v
u
′
v
2
−
u
v
′
v
2
=
v
u
′
−
u
v
′
v
2
{\displaystyle y'={\frac {u'}{v}}-{\frac {uv'}{v^{2}}}={\frac {vu'}{v^{2}}}-{\frac {uv'}{v^{2}}}={\frac {vu'-uv'}{v^{2}}}}
Derivatives of trigonometric functions [ edit | edit source ]
y
=
sin
(
x
)
{\displaystyle y=\sin(x)}
y
+
Δ
y
=
sin
(
x
+
Δ
x
)
=
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
{\displaystyle y+\Delta y=\sin(x+\Delta x)=\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x)}
Δ
y
=
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
−
sin
(
x
)
{\displaystyle \Delta y=\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x)-\sin(x)}
Δ
y
Δ
x
=
sin
(
x
)
{\displaystyle {\frac {\Delta y}{\Delta x}}=\sin(x)}
⋅
cos
(
Δ
x
)
−
1
Δ
x
+
cos
(
x
)
{\displaystyle {\frac {\cos(\Delta x)-1}{\Delta x}}+\cos(x)}
⋅
sin
(
Δ
x
)
Δ
x
{\displaystyle {\frac {\sin(\Delta x)}{\Delta x}}}
The value
cos
(
Δ
x
)
−
1
Δ
x
{\displaystyle {\frac {\cos(\Delta x)-1}{\Delta x}}}
>>> # python code
>>> [ ( math . cos ( Δx ) - 1 ) / Δx for Δx in ( .1 , .01 , .001 , .0001 , .0000_1 , .0000_01 , .0000_001 , .0000_0001 , .0000_0000_1 ) ]
[ - 0.049958347219742905 , - 0.004999958333473664 , - 0.0004999999583255033 ,
- 4.999999969612645e-05 , - 5.000000413701855e-06 , - 5.000444502911705e-07 ,
- 4.9960036108132044e-08 , 0.0 , 0.0 ]
>>>
lim
Δ
x
→
0
cos
(
Δ
x
)
−
1
Δ
x
=
lim
Δ
x
→
0
−
sin
(
Δ
x
)
1
=
0
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {\cos(\Delta x)-1}{\Delta x}}=\lim _{\Delta x\rightarrow 0}{\frac {-\sin(\Delta x)}{1}}=0}
The value
sin
(
Δ
x
)
Δ
x
{\displaystyle {\frac {\sin(\Delta x)}{\Delta x}}}
>>> # python code
>>> [ math . sin ( Δx ) / Δx for Δx in ( .1 , .01 , .001 , .0001 , .0000_1 , .0000_01 , .0000_001 , .0000_0001 , .0000_0000_1 ) ]
[ 0.9983341664682815 , 0.9999833334166665 , 0.9999998333333416 ,
0.9999999983333334 , 0.9999999999833332 , 0.9999999999998334 ,
0.9999999999999983 , 1.0 , 1.0 ]
>>>
lim
Δ
x
→
0
sin
(
Δ
x
)
Δ
x
=
lim
Δ
x
→
0
cos
(
Δ
x
)
1
=
1
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {\sin(\Delta x)}{\Delta x}}=\lim _{\Delta x\rightarrow 0}{\frac {\cos(\Delta x)}{1}}=1}
d
y
d
x
=
sin
(
x
)
{\displaystyle {\frac {dy}{dx}}=\sin(x)}
⋅
0
+
cos
(
x
)
{\displaystyle 0+\cos(x)}
⋅
1
=
cos
(
x
)
{\displaystyle 1=\cos(x)}
y
=
cos
(
x
)
=
1
−
sin
2
(
x
)
{\displaystyle y=\cos(x)={\sqrt {1-\sin ^{2}(x)}}}
y
2
=
1
−
sin
(
x
)
{\displaystyle y^{2}=1-\sin(x)}
⋅
sin
(
x
)
{\displaystyle \sin(x)}
Differentiate both sides:
2
y
{\displaystyle 2y}
⋅
d
y
d
x
=
−
(
2
sin
(
x
)
cos
(
x
)
)
{\displaystyle {\frac {dy}{dx}}=-(2\sin(x)\cos(x))}
d
y
d
x
=
−
sin
(
x
)
cos
(
x
)
cos
(
x
)
=
−
sin
(
x
)
.
{\displaystyle {\frac {dy}{dx}}=-{\frac {\sin(x)\cos(x)}{\cos(x)}}=-\sin(x).}
y
=
tan
(
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle y=\tan(x)={\frac {\sin(x)}{\cos(x)}}}
y
cos
(
x
)
=
sin
(
x
)
{\displaystyle y\cos(x)=\sin(x)}
Differentiate both sides:
y
(
−
sin
(
x
)
)
+
cos
(
x
)
{\displaystyle y(-\sin(x))+\cos(x)}
⋅
d
y
d
x
=
cos
(
x
)
{\displaystyle {\frac {dy}{dx}}=\cos(x)}
cos
(
x
)
{\displaystyle \cos(x)}
⋅
d
y
d
x
=
cos
(
x
)
+
y
sin
(
x
)
{\displaystyle {\frac {dy}{dx}}=\cos(x)+y\sin(x)}
d
y
d
x
=
cos
(
x
)
+
tan
(
x
)
sin
(
x
)
cos
(
x
)
=
1
+
tan
2
(
x
)
=
sec
2
(
x
)
{\displaystyle {\frac {dy}{dx}}={\frac {\cos(x)+\tan(x)\sin(x)}{\cos(x)}}=1+\tan ^{2}(x)=\sec ^{2}(x)}
Derivatives of inverse trigonometric functions [ edit | edit source ] Derivatives of logarithmic functions [ edit | edit source ] a^x or
a
x
{\displaystyle a^{x}}
[ edit | edit source ]
y
=
a
x
{\displaystyle y=a^{x}}
y
+
Δ
y
=
a
x
+
Δ
x
=
a
x
a
Δ
x
{\displaystyle y+\Delta y=a^{x+\Delta x}=a^{x}a^{\Delta x}}
Δ
y
=
a
x
a
Δ
x
−
a
x
=
a
x
(
a
Δ
x
−
1
)
{\displaystyle \Delta y=a^{x}a^{\Delta x}-a^{x}=a^{x}(a^{\Delta x}-1)}
Δ
y
Δ
x
=
a
x
⋅
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}=a^{x}\cdot {\frac {a^{\Delta x}-1}{\Delta x}}}
Consider the value
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {a^{\Delta x}-1}{\Delta x}}}
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}}
d
(
a
x
)
d
x
{\displaystyle {\frac {d(a^{x})}{dx}}}
a
1
2
=
a
{\displaystyle a^{\frac {1}{2}}={\sqrt {a}}}
(
a
1
2
)
1
2
=
a
{\displaystyle (a^{\frac {1}{2}})^{\frac {1}{2}}={\sqrt {\sqrt {a}}}}
a
1
2
3
=
a
{\displaystyle a^{\frac {1}{2^{3}}}={\sqrt {\sqrt {\sqrt {a}}}}}
a
1
2
n
=
a
{\displaystyle a^{\frac {1}{2^{n}}}={\sqrt {a}}}
n
{\displaystyle n}
We will look at the expression
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {a^{\Delta x}-1}{\Delta x}}}
a
{\displaystyle a}
Δ
x
=
1
2
70
=
8.5
e
−
22
{\displaystyle \Delta x={\frac {1}{2^{70}}}=8.5e-22}
(
a
1
/
2
70
−
1
)
∗
(
2
70
)
{\displaystyle (a^{1/2^{70}}-1)*(2^{70})}
a
1
/
2
70
{\displaystyle a^{1/2^{70}}}
a
70
{\displaystyle {\sqrt[{70}]{a}}}
a
{\displaystyle {\sqrt {a}}}
70
{\displaystyle 70}
sqrt() can be written so
that it does not depend on logarithmic or exponential operations.
>>> # python code
>>> N = Decimal ( 2 )
>>> v2 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v2
Decimal ( '0.69314718055994530941743560122437474084363865015406919942144' )
>>>
>>> N = Decimal ( 8 )
>>> v8 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v8
Decimal ( '2.07944154167983592825352768227031325913255072732801782513664' )
>>>
>>> N = Decimal ( 32 )
>>> v32 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v32
Decimal ( '3.46573590279972654709124760144583715956091114572543812435968' )
>>>
>>> N = Decimal ( 128 )
>>> v128 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v128
Decimal ( '4.85203026391961716593059535875094644210510807293198187102208' )
>>>
Compare the values v8, v32, v128 with v2:
>>> v8 / v2 ; v32 / v2 ; v128 / v2
Decimal ( '3.00000000000000000000176135549769209744528640235368520610520' )
Decimal ( '5.00000000000000000000587118499230699148996545343403093128046' )
Decimal ( '7.00000000000000000001232948848384468209997247971055570994695' )
>>>
We know that
8
=
2
3
;
32
=
2
5
;
128
=
2
7
{\displaystyle 8=2^{3};\ 32=2^{5};\ 128=2^{7}}
v2, v8, v32, v128 are behaving like logarithms.
In fact
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}}
a
{\displaystyle a}
ln
(
a
)
.
{\displaystyle \ln(a).}
Figure 5: Graphs of
a
x
−
1
x
{\displaystyle {\frac {a^{x}-1}{x}}}
for a =
2
,
8
,
32
,
128
{\displaystyle 2,\ 8,\ 32,\ 128}
. Figure 5 contains graphs of
a
x
−
1
x
{\displaystyle {\frac {a^{x}-1}{x}}}
a
=
2
,
8
,
32
,
128
{\displaystyle a=2,\ 8,\ 32,\ 128}
e
x
−
1
x
{\displaystyle {\frac {e^{x}-1}{x}}}
All values of
x
{\displaystyle x}
x
=
0.
{\displaystyle x=0.}
The correct value of
ln
(
2
)
{\displaystyle \ln(2)}
>>> Decimal ( 2 ) . ln ()
Decimal ( '0.693147180559945309417232121458176568075500134360255254120680' )
>>>
Our calculation produced:
Decimal ( '0.69314718055994530941743560122437474084363865015406919942144' )
accurate to 21 places of decimals, not bad for one line of simple python code using high-school math.
This method for calculation of
ln
(
a
)
{\displaystyle \ln(a)}
sqrt() is available.
Programming language python interprets expression a**b as
a
b
.
{\displaystyle a^{b}.}
ln
(
2
)
{\displaystyle \ln(2)}
# python code
>>> getcontext () . prec
101 # Precision of 101.
>>> dx = Decimal ( '1E-50' )
>>> ( 2 ** dx - 1 ) / dx
Decimal ( '0.69314718055994530941723212145817656807550013436026' ) # ln(2)
>>> ( 2 ** ( - dx ) - 1 ) / ( - dx )
Decimal ( '0.693147180559945309417232121458176568075500134360253' ) # ln(2)
>>>
When
y
=
a
x
,
d
y
d
x
=
a
x
⋅
ln
(
a
)
.
{\displaystyle y=a^{x},{\frac {dy}{dx}}=a^{x}\cdot \ln(a).}
The base of natural logarithms is the value of
a
{\displaystyle a}
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
=
1.
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}=1.}
This value of
a
{\displaystyle a}
e
,
=
2.718281828459045235360287471352662497757247093699959574.....
{\displaystyle e,=2.718281828459045235360287471352662497757247093699959574.....}
>>> # python code
>>> N = e = Decimal ( 1 ) . exp (); N
Decimal ( '2.71828182845904523536028747135266249775724709369995957496697' )
>>> ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 )
Decimal ( '1.0000_0000_0000_0000_0000_0423_51647362715016770651530003719651328' )
>>> # ln(e) = 1. Our calculation of ln(e) is accurate to 21 places of decimals.
When
a
=
e
,
d
(
e
x
)
d
x
=
e
x
{\displaystyle a=e,{\frac {d(e^{x})}{dx}}=e^{x}}
⋅
ln
(
e
)
=
e
x
{\displaystyle \ln(e)=e^{x}}
⋅
1
=
e
x
.
{\displaystyle 1=e^{x}.}
y
=
l
n
(
x
)
{\displaystyle y=ln(x)}
x
=
e
y
{\displaystyle x=e^{y}}
d
x
d
y
=
e
y
{\displaystyle {\frac {dx}{dy}}=e^{y}}
d
y
d
x
=
1
e
y
=
1
x
{\displaystyle {\frac {dy}{dx}}={\frac {1}{e^{y}}}={\frac {1}{x}}}
y
=
ln
(
a
x
)
{\displaystyle y=\ln(ax)}
d
y
d
x
=
d
d
x
ln
(
a
x
)
=
d
d
x
(
ln
(
a
)
+
ln
(
x
)
)
=
d
d
x
ln
(
a
)
+
d
d
x
ln
(
x
)
=
1
x
{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\ln(ax)={\frac {d}{dx}}(\ln(a)+\ln(x))={\frac {d}{dx}}\ln(a)+{\frac {d}{dx}}\ln(x)={\frac {1}{x}}}
y
=
ln
(
x
a
)
{\displaystyle y=\ln(x^{a})}
d
y
d
x
=
d
d
x
ln
(
x
a
)
=
d
d
x
(
a
⋅
ln
(
x
)
)
=
a
⋅
d
d
x
ln
(
x
)
=
a
x
{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\ln(x^{a})={\frac {d}{dx}}(a\cdot \ln(x))=a\cdot {\frac {d}{dx}}\ln(x)={\frac {a}{x}}}
y
=
x
m
n
{\displaystyle y=x^{\frac {m}{n}}}
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
y
n
=
x
m
{\displaystyle y^{n}=x^{m}}
n
⋅
ln
(
y
)
=
m
⋅
ln
(
x
)
{\displaystyle n\cdot \ln(y)=m\cdot \ln(x)}
n
⋅
1
y
⋅
d
y
d
x
=
m
⋅
1
x
{\displaystyle n\cdot {\frac {1}{y}}\cdot {\frac {dy}{dx}}=m\cdot {\frac {1}{x}}}
d
y
d
x
=
m
⋅
1
x
⋅
y
n
{\displaystyle {\frac {dy}{dx}}=m\cdot {\frac {1}{x}}\cdot {\frac {y}{n}}}
=
m
n
⋅
x
m
n
x
{\displaystyle ={\frac {m}{n}}\cdot {\frac {x^{\frac {m}{n}}}{x}}}
=
m
n
⋅
x
m
n
−
1
{\displaystyle ={\frac {m}{n}}\cdot x^{{\frac {m}{n}}-1}}
Careful manipulation of logarithms converts exponents into simple constants.
Used where
y
=
G
(
H
(
I
(
x
)
)
)
{\displaystyle y=G(H(I(x)))}
y
=
cos
(
2
x
)
{\displaystyle y=\cos(2x)}
Let
y
=
cos
(
u
)
{\displaystyle y=\cos(u)}
u
=
2
x
{\displaystyle u=2x}
d
y
d
x
=
d
y
d
u
⋅
d
u
d
x
=
−
sin
(
u
)
⋅
2
=
−
2
sin
(
2
x
)
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}=&{\frac {dy}{du}}\cdot {\frac {du}{dx}}\\=&-\sin(u)\cdot 2\\=&-2\sin(2x)\\\end{aligned}}}
y
=
sin
2
(
ln
(
5
x
)
)
{\displaystyle y=\sin ^{2}(\ln(5x))}
Let
y
=
u
2
{\displaystyle y=u^{2}}
u
=
sin
(
v
)
;
v
=
ln
(
w
)
;
w
=
5
x
{\displaystyle u=\sin(v);\ v=\ln(w);\ w=5x}
d
y
d
x
=
d
y
d
u
⋅
d
u
d
v
⋅
d
v
d
w
⋅
d
w
d
x
=
2
u
⋅
cos
(
v
)
⋅
1
w
⋅
5
=
2
sin
(
v
)
⋅
cos
(
ln
(
w
)
)
⋅
1
5
x
⋅
5
=
2
sin
(
ln
(
w
)
)
⋅
cos
(
ln
(
5
x
)
)
⋅
1
x
=
2
sin
(
ln
(
5
x
)
)
⋅
cos
(
ln
(
5
x
)
)
⋅
1
x
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}=&{\frac {dy}{du}}\cdot {\frac {du}{dv}}\cdot {\frac {dv}{dw}}\cdot {\frac {dw}{dx}}\\=&2u\cdot \cos(v)\cdot {\frac {1}{w}}\cdot 5\\=&2\sin(v)\cdot \cos(\ln(w))\cdot {\frac {1}{5x}}\cdot 5\\=&2\sin(\ln(w))\cdot \cos(\ln(5x))\cdot {\frac {1}{x}}\\=&2\sin(\ln(5x))\cdot \cos(\ln(5x))\cdot {\frac {1}{x}}\\\end{aligned}}}
Applications of the Derivative [ edit | edit source ] The first derivative of
f
(
x
)
=
y
′
{\displaystyle f(x)=y'}
f
′
(
x
)
.
{\displaystyle f'(x).}
f
′
(
x
)
{\displaystyle f'(x)}
x
1
{\displaystyle x_{1}}
f
(
x
)
{\displaystyle f(x)}
x
1
.
{\displaystyle x_{1}.}
f
′
(
x
1
)
{\displaystyle f'(x_{1})}
f
(
x
)
{\displaystyle f(x)}
x
=
x
1
.
{\displaystyle x=x_{1}.}
Figure 1: Diagram illustrating relationship between
f
(
x
)
=
x
2
−
x
−
2
{\displaystyle f(x)=x^{2}-x-2}
f
′
(
x
)
=
2
x
−
1.
{\displaystyle f'(x)=2x-1.}
x
=
0.5
,
{\displaystyle x=0.5,}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
x
=
0
,
{\displaystyle x=0,}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
−
1.
{\displaystyle f(x)=-1.}
x
=
1
,
{\displaystyle x=1,}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
1.
{\displaystyle f(x)=1.}
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
In the example to the right,
y
=
f
(
x
)
=
x
2
−
x
−
2
{\displaystyle y=f(x)=x^{2}-x-2}
y
′
=
f
′
(
x
)
=
2
x
−
1.
{\displaystyle y'=f'(x)=2x-1.}
Of special interest is the point at which
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
f
′
(
x
)
=
0
,
x
=
0.5
{\displaystyle f'(x)=0,\ x=0.5}
f
(
x
)
=
0.5
2
−
0.5
−
2
=
−
2.25.
{\displaystyle f(x)=0.5^{2}-0.5-2=-2.25.}
The point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
f
(
x
)
.
{\displaystyle f(x).}
y
′
{\displaystyle y'}
y
′
=
0
,
f
(
x
)
{\displaystyle y'=0,\ f(x)}
The value of
y
{\displaystyle y}
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
y
{\displaystyle y}
(
0
,
−
2
)
,
(
1
,
−
2
)
.
{\displaystyle (0,-2),\ (1,-2).}
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
f
(
x
)
.
{\displaystyle f(x).}
In this curve
y
=
x
2
−
x
−
2
,
{\displaystyle y=x^{2}-x-2,}
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
Figure 1: Diagram illustrating relationship between
f
(
x
)
=
2
x
3
+
3
x
2
−
12
x
−
8
8
{\displaystyle f(x)={\frac {2x^{3}+3x^{2}-12x-8}{8}}}
f
′
(
x
)
=
6
x
2
+
6
x
−
12
8
.
{\displaystyle f'(x)={\frac {6x^{2}+6x-12}{8}}.}
x
=
−
2
{\displaystyle x=-2}
x
=
1
,
{\displaystyle x=1,}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
x
=
−
3
,
{\displaystyle x=-3,}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
3.
{\displaystyle f(x)=3.}
x
=
−
1
,
{\displaystyle x=-1,}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
−
1.5.
{\displaystyle f(x)=-1.5.}
x
=
2
,
{\displaystyle x=2,}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
3.
{\displaystyle f(x)=3.}
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
In the example to the right,
y
=
f
(
x
)
=
2
x
3
+
3
x
2
−
12
x
−
8
8
{\displaystyle y=f(x)={\frac {2x^{3}+3x^{2}-12x-8}{8}}}
y
′
=
f
′
(
x
)
=
6
x
2
+
6
x
−
12
8
.
{\displaystyle y'=f'(x)={\frac {6x^{2}+6x-12}{8}}.}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
f
′
(
x
)
=
0
,
x
=
−
2
{\displaystyle f'(x)=0,\ x=-2}
f
(
x
)
=
1.5
{\displaystyle f(x)=1.5}
When
f
′
(
x
)
=
0
,
x
=
1
{\displaystyle f'(x)=0,\ x=1}
f
(
x
)
=
−
1
7
8
.
{\displaystyle f(x)=-1{\frac {7}{8}}.}
The points
(
−
2
,
1.5
)
,
(
1
,
−
1
7
8
)
{\displaystyle (-2,1.5),\ (1,-1{\frac {7}{8}})}
f
(
x
)
.
{\displaystyle f(x).}
y
′
{\displaystyle y'}
y
′
=
0
,
f
(
x
)
{\displaystyle y'=0,\ f(x)}
f
(
x
)
{\displaystyle f(x)}
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
f
(
x
)
{\displaystyle f(x)}
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
f
(
x
)
{\displaystyle f(x)}
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
f
(
x
)
{\displaystyle f(x)}
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
Figure 1(c): Plan of county road between Town A and Town B to be constructed so that cost is minimum. Town B is 40 miles East and 50 miles North of Town A. The county is going to construct a road from Town A to Town B.
Adjacent to Town A the cost to build a road is $500k/mile.
Adjacent to Town B the cost to build a road is $200k/mile.
The dividing line runs East-West 30 miles North of Town A.
Calculate the position of point C so that the cost of the road from Town A to Town B is minimum.
Let point
C
=
(
x
,
30
)
.
{\displaystyle C=(x,30).}
Then distance from Town A to point
C
=
x
2
+
30
2
=
x
2
+
900
.
{\displaystyle C={\sqrt {x^{2}+30^{2}}}={\sqrt {x^{2}+900}}.}
Distance from Town B to point
C
=
(
40
−
x
)
2
+
20
2
{\displaystyle C={\sqrt {(40-x)^{2}+20^{2}}}}
=
1600
−
80
x
+
x
2
+
400
{\displaystyle ={\sqrt {1600-80x+x^{2}+400}}}
=
2000
−
80
x
+
x
2
.
{\displaystyle ={\sqrt {2000-80x+x^{2}}}.}
x
=
10.52322823517
…
{\displaystyle x=10.52322823517\dots }
Figure 1(e): Sheet of cardboard to be cut and folded to make box of maximum possible volume. A piece of cardboard of length
4
f
t
{\displaystyle 4\ ft}
3
f
t
{\displaystyle 3\ ft}
What is the height of the box?
Figure 1: Graph of ellipse showing semi-chord through center.
t
{\displaystyle t}
=
2
t
.
{\displaystyle =2t.}
t
{\displaystyle t}
=
2
t
.
{\displaystyle =2t.}
An ellipse with center at origin has equation:
A
x
2
+
B
x
y
+
C
y
2
+
F
=
0
…
(
1
)
{\displaystyle Ax^{2}+Bxy+Cy^{2}+F=0\ \dots \ (1)}
Given values
A
,
B
,
C
,
F
{\displaystyle A,B,C,F}
In Figure 1
t
{\displaystyle t}
θ
{\displaystyle \theta }
X
{\displaystyle X}
t
.
{\displaystyle t.}
Aim of this section is to calculate
θ
{\displaystyle \theta }
t
{\displaystyle t}
2
t
.
{\displaystyle 2t.}
Let
c
=
cos
(
θ
)
{\displaystyle c=\cos(\theta )}
s
=
sin
(
θ
)
.
{\displaystyle s=\sin(\theta ).}
Then
x
=
t
cos
(
θ
)
=
t
c
,
{\displaystyle x=t\cos(\theta )=tc,}
y
=
t
sin
(
θ
)
=
t
s
.
{\displaystyle y=t\sin(\theta )=ts.}
Substitute these values in
(
1
)
:
{\displaystyle (1):}
A
t
t
c
c
+
B
t
t
c
s
+
C
t
t
s
s
+
F
=
0
…
(
2
)
{\displaystyle Attcc+Bttcs+Cttss+F=0\ \dots \ (2)}
Calculate
t
′
=
d
t
d
θ
{\displaystyle t'={\frac {dt}{d\theta }}}
A
(
t
t
(
−
2
c
s
)
+
c
c
2
t
t
′
)
+
B
(
t
t
(
c
c
−
s
s
)
+
c
s
2
t
t
′
)
+
C
(
t
t
(
2
s
c
)
+
s
s
2
t
t
′
)
=
0
{\displaystyle A(tt(-2cs)+cc2tt')+B(tt(cc-ss)+cs2tt')+C(tt(2sc)+ss2tt')=0}
A
t
t
(
−
2
c
s
)
+
A
c
c
2
t
t
′
+
B
t
t
(
c
c
−
s
s
)
+
B
c
s
2
t
t
′
+
C
t
t
(
2
s
c
)
+
C
s
s
2
t
t
′
=
0
{\displaystyle Att(-2cs)+Acc2tt'+Btt(cc-ss)+Bcs2tt'+Ctt(2sc)+Css2tt'=0}
−
2
A
t
t
c
s
+
A
c
c
2
t
t
′
+
B
t
t
c
c
−
B
t
t
s
s
+
B
c
s
2
t
t
′
+
2
C
t
t
s
c
+
2
C
s
s
t
t
′
=
0
{\displaystyle -2Attcs+Acc2tt'+Bttcc-Bttss+Bcs2tt'+2Cttsc+2Csstt'=0}
+
A
c
c
2
t
t
′
+
2
B
c
s
t
t
′
+
2
C
s
s
t
t
′
−
2
A
t
t
c
s
+
B
t
t
c
c
−
B
t
t
s
s
+
2
C
t
t
s
c
=
0
{\displaystyle +Acc2tt'+2Bcstt'+2Csstt'-2Attcs+Bttcc-Bttss+2Cttsc=0}
+
t
′
(
A
c
c
2
t
+
2
B
c
s
t
+
2
C
s
s
t
)
=
+
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
{\displaystyle +t'(Acc2t+2Bcst+2Csst)=+2Attcs-Bttcc+Bttss-2Cttsc}
t
′
=
+
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
(
A
c
c
2
t
+
2
B
c
s
t
+
2
C
s
s
t
)
{\displaystyle t'={\frac {+2Attcs-Bttcc+Bttss-2Cttsc}{(Acc2t+2Bcst+2Csst)}}}
For maximum or minimum
t
:
{\displaystyle t:}
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
=
0
{\displaystyle 2Attcs-Bttcc+Bttss-2Cttsc=0}
2
A
c
s
−
B
c
c
+
B
s
s
−
2
C
s
c
=
0
{\displaystyle 2Acs-Bcc+Bss-2Csc=0}
2
A
c
s
−
2
C
s
c
=
B
c
c
−
B
s
s
{\displaystyle 2Acs-2Csc=Bcc-Bss}
Square both sides, substitute
1
−
s
s
{\displaystyle 1-ss}
c
c
{\displaystyle cc}
a
S
2
+
b
S
+
c
=
0
…
(
3
)
{\displaystyle aS^{2}+bS+c=0\ \dots \ (3)}
S
=
sin
2
(
θ
)
{\displaystyle S=\sin ^{2}(\theta )}
a
=
(
+
4
A
A
−
8
A
C
+
4
B
B
+
4
C
C
)
{\displaystyle a=(+4AA-8AC+4BB+4CC)}
b
=
−
a
{\displaystyle b=-a}
c
=
B
B
{\displaystyle c=BB}
Let equation of ellipse be:
55
x
2
−
24
x
y
+
48
y
2
−
2496
=
0
{\displaystyle 55x^{2}-24xy+48y^{2}-2496=0}
# python code
>>> A , B , C = 55 , - 24 , 48
>>> a = ( + 4 * A * A - 8 * A * C + 4 * B * B + 4 * C * C ); a
2500
>>> b = - a ; b
- 2500
>>> c = B * B ; c
576
>>>
>>> a , b , c = [ v / 4 for v in ( a , b , c )] ; a , b , c
( 625.0 , - 625.0 , 144.0 )
>>> S = .36
>>> a * S * S + b * S + c
0.0
>>> S = .64
>>> a * S * S + b * S + c
0.0
>>>
The solutions of quadratic equation
(
3
)
{\displaystyle (3)}
.36
{\displaystyle .36}
.64
{\displaystyle .64}
Therefore
sin
(
θ
)
=
±
0.6
{\displaystyle \sin(\theta )=\pm 0.6}
sin
(
θ
)
=
±
0.8
{\displaystyle \sin(\theta )=\pm 0.8}
From
(
2
)
{\displaystyle (2)}
t
=
−
F
A
c
c
+
B
c
s
+
C
s
s
{\displaystyle t={\sqrt {\frac {-F}{Acc+Bcs+Css}}}}
# python code
A , B , C , F = 55 , - 24 , 48 , - 2496
t1 = ( 0.6 , 0.8 )
dict1 = dict ()
for v1 in ( t1 , t1 [:: - 1 ]) :
c1 , s1 = v1
for c in ( c1 , - c1 ) :
for s in ( s1 , - s1 ) :
t = ( - F / ( A * c * c + B * c * s + C * s * s )) ** .5
if t in dict1 : dict1 [ t ] += (( c , s ),)
else : dict1 [ t ] = (( c , s ),)
L1 = [ ( v , dict1 [ v ]) for v in sorted ([ v for v in dict1 ]) ]
for v in L1 : print ( v )
(6.244997998398398, ((0.8, -0.6), (-0.8, 0.6)))
(6.34287855135306, ((0.6, -0.8), (-0.6, 0.8)))
(7.806247497997998, ((0.8, 0.6), (-0.8, -0.6)))
(7.999999999999999, ((0.6, 0.8), (-0.6, -0.8)))
Minimum value of
t
=
6.244997998398398.
{\displaystyle t=6.244997998398398.}
=
2
∗
6.244997998398398
{\displaystyle =2*6.244997998398398}
Maximum value of
t
=
7.999999999999999.
{\displaystyle t=7.999999999999999.}
=
8
∗
2
{\displaystyle =8*2}
Figure 3: Curves and values associated with car jack.
x
=
1
,
d
y
d
x
=
0.1
…
{\displaystyle x=1,\ {\frac {dy}{dx}}=0.1\dots }
θ
=
45
∘
,
d
y
d
x
=
1
{\displaystyle \theta =45^{\circ },\ {\frac {dy}{dx}}=1}
x
=
9
,
d
y
d
x
=
2.06
…
{\displaystyle x=9,\ {\frac {dy}{dx}}=2.06\dots }
At what rate is point
A
{\displaystyle A}
(a) when
x
=
9
{\displaystyle x=9}
(b) when
x
=
1
{\displaystyle x=1}
(c) when
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
We have to calculate
d
y
d
t
{\displaystyle {\frac {dy}{dt}}}
d
x
d
t
{\displaystyle {\frac {dx}{dt}}}
d
y
d
t
=
d
y
d
x
⋅
d
x
d
t
{\displaystyle {\frac {dy}{dt}}={\frac {dy}{dx}}\cdot {\frac {dx}{dt}}}
x
2
+
y
2
=
10
2
{\displaystyle x^{2}+y^{2}=10^{2}}
y
=
100
−
x
2
{\displaystyle y={\sqrt {100-x^{2}}}}
y
2
=
100
−
x
2
{\displaystyle y^{2}=100-x^{2}}
2
y
⋅
d
y
d
x
=
−
2
x
{\displaystyle 2y\cdot {\frac {dy}{dx}}=-2x}
d
y
d
x
=
−
2
x
2
y
=
−
x
100
−
x
2
{\displaystyle {\frac {dy}{dx}}={\frac {-2x}{2y}}={\frac {-x}{\sqrt {100-x^{2}}}}}
For convenience we'll use the negative value of the square root and say that
d
y
d
x
=
x
100
−
x
2
.
{\displaystyle {\frac {dy}{dx}}={\frac {x}{\sqrt {100-x^{2}}}}.}
Relative to line
B
C
:
{\displaystyle BC:}
When
x
=
9
,
d
y
d
x
=
2.06
,
d
y
d
t
=
2.06
⋅
5
=
10.3
{\displaystyle x=9,\ {\frac {dy}{dx}}=2.06,\ {\frac {dy}{dt}}=2.06\cdot 5=10.3}
/
{\displaystyle /}
When
x
=
1
,
d
y
d
x
=
0.1
,
d
y
d
t
=
0.1
⋅
5
=
0.5
{\displaystyle x=1,\ {\frac {dy}{dx}}=0.1,\ {\frac {dy}{dt}}=0.1\cdot 5=0.5}
/
{\displaystyle /}
When
θ
=
45
∘
,
x
=
10
⋅
cos
45
∘
=
7.071
{\displaystyle \theta =45^{\circ },x=10\cdot \cos 45^{\circ }=7.071}
d
y
d
t
=
1
⋅
5
=
5
{\displaystyle {\frac {dy}{dt}}=1\cdot 5=5}
/
{\displaystyle /}
This example highlights the mechanical advantage of this simple but effective tool. When the top of the jack is low, it moves quickly.
As the jack takes more and more weight, the top of the jack moves more slowly.
Figure 4: Graph of
a
{\displaystyle a}
and
d
a
d
x
{\displaystyle {\frac {da}{dx}}}
.
Δ
A
B
C
{\displaystyle \Delta ABC}
b
=
h
=
50
=
10
cos
(
45
∘
)
{\displaystyle b=h={\sqrt {50}}=10\cos(45^{\circ })}
At what rate is the area of
Δ
A
B
C
{\displaystyle \Delta ABC}
(i)
x
=
9
{\displaystyle x=9}
(ii)
x
=
1
{\displaystyle x=1}
(iii)
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
d
a
d
t
=
d
a
d
x
⋅
d
x
d
t
{\displaystyle {\frac {da}{dt}}={\frac {da}{dx}}\cdot {\frac {dx}{dt}}}
a
{\displaystyle a}
Δ
A
B
C
{\displaystyle \Delta ABC}
d
x
d
t
=
5
{\displaystyle {\frac {dx}{dt}}=5}
/
{\displaystyle /}
a
=
b
⋅
h
2
=
x
⋅
y
2
=
x
100
−
x
2
2
{\displaystyle a={\frac {b\cdot h}{2}}={\frac {x\cdot y}{2}}={\frac {x{\sqrt {100-x^{2}}}}{2}}}
Calculate
d
a
d
x
:
{\displaystyle {\frac {da}{dx}}:}
2
a
=
x
100
−
x
2
{\displaystyle 2a=x{\sqrt {100-x^{2}}}}
4
a
2
=
x
2
(
100
−
x
2
)
=
100
x
2
−
x
4
{\displaystyle 4a^{2}=x^{2}(100-x^{2})=100x^{2}-x^{4}}
8
a
⋅
d
a
d
x
=
200
x
−
4
x
3
{\displaystyle 8a\cdot {\frac {da}{dx}}=200x-4x^{3}}
d
a
d
x
=
200
x
−
4
x
3
8
a
=
50
x
−
x
3
2
a
=
50
x
−
x
3
x
100
−
x
2
{\displaystyle {\frac {da}{dx}}={\frac {200x-4x^{3}}{8a}}={\frac {50x-x^{3}}{2a}}={\frac {50x-x^{3}}{x{\sqrt {100-x^{2}}}}}}
=
50
−
x
2
100
−
x
2
{\displaystyle ={\frac {50-x^{2}}{\sqrt {100-x^{2}}}}}
When
x
=
9
,
d
a
d
x
=
7.11
…
{\displaystyle x=9,\ {\frac {da}{dx}}=7.11\dots }
Δ
A
B
C
{\displaystyle \Delta ABC}
(
5
⋅
7.11
…
)
{\displaystyle (5\cdot 7.11\dots )}
When
x
=
1
,
d
a
d
x
=
−
4.92
…
{\displaystyle x=1,\ {\frac {da}{dx}}=-4.92\dots }
Δ
A
B
C
{\displaystyle \Delta ABC}
(
5
⋅
4.92
…
)
{\displaystyle (5\cdot 4.92\dots )}
When
θ
=
45
∘
,
h
=
b
=
x
=
y
=
10
cos
(
45
∘
)
=
10
⋅
2
2
=
5
2
,
{\displaystyle \theta =45^{\circ },\ h=b=x=y=10\cos(45^{\circ })=10\cdot {\frac {\sqrt {2}}{2}}=5{\sqrt {2}},\ }
d
a
d
x
=
0
,
a
{\displaystyle {\frac {da}{dx}}=0,\ a}
25
{\displaystyle 25}
d
a
d
t
=
0.
{\displaystyle {\frac {da}{dt}}=0.}
{\displaystyle }
Figure 5: Image of analog clock showing minute and hour hands at
3
{\displaystyle 3}
o'clock. An old fashion analog clock with American style face (12 hours) keeps accurate time.
The length of the minute hand is
4
{\displaystyle 4}
3
{\displaystyle 3}
At what rate is the tip of the minute hand approaching the tip of the hour hand at 3 o'clock?
Let
a
{\displaystyle a}
The task is to calculate
d
a
d
t
.
{\displaystyle {\frac {da}{dt}}.}
d
a
d
t
=
d
a
d
θ
⋅
d
θ
d
t
{\displaystyle {\frac {da}{dt}}={\frac {da}{d\theta }}\cdot {\frac {d\theta }{dt}}}
θ
{\displaystyle \theta }
a
{\displaystyle a}
Calculating
d
θ
d
t
:
{\displaystyle {\frac {d\theta }{dt}}:}
Angular velocity of minute hand
=
2
π
1
{\displaystyle ={\frac {2\pi }{1}}}
Angular velocity of hour hand
=
2
π
12
=
π
6
{\displaystyle ={\frac {2\pi }{12}}={\frac {\pi }{6}}}
d
θ
d
t
=
{\displaystyle {\frac {d\theta }{dt}}=}
=
2
π
−
π
6
=
11
π
6
{\displaystyle =2\pi -{\frac {\pi }{6}}={\frac {11\pi }{6}}}
Calculating
d
a
d
θ
:
{\displaystyle {\frac {da}{d\theta }}:}
a
2
=
3
2
+
4
2
−
2
⋅
3
⋅
4
⋅
cos
θ
=
25
−
24
cos
θ
.
{\displaystyle a^{2}=3^{2}+4^{2}-2\cdot 3\cdot 4\cdot \cos \theta =25-24\cos \theta .}
2
a
⋅
d
a
d
θ
=
24
sin
θ
.
{\displaystyle 2a\cdot {\frac {da}{d\theta }}=24\sin \theta .}
d
a
d
θ
=
24
sin
θ
2
a
=
24
sin
90
∘
2
⋅
5
=
24
10
=
2.4.
{\displaystyle {\frac {da}{d\theta }}={\frac {24\sin \theta }{2a}}={\frac {24\sin 90^{\circ }}{2\cdot 5}}={\frac {24}{10}}=2.4.}
d
a
d
t
=
2.4
⋅
11
π
6
=
4.4
π
{\displaystyle {\frac {da}{dt}}=2.4\cdot {\frac {11\pi }{6}}=4.4\pi }
Figure 2: Diagram showing position of piston as a function of rotation of crankshaft in reciprocating engine.
8
{\displaystyle 8}
18
{\displaystyle 18}
18
≥
y
≥
8
{\displaystyle 18\geq y\geq 8}
a
2
=
y
2
+
c
2
−
2
y
c
cos
(
x
)
{\displaystyle a^{2}=y^{2}+c^{2}-2yc\cos(x)}
y
2
−
2
c
cos
(
x
)
y
+
c
2
−
a
2
=
0
{\displaystyle y^{2}-2c\cos(x)y+c^{2}-a^{2}=0}
y
2
−
2
(
5
)
cos
(
x
)
y
+
(
5
)
2
−
(
13
)
2
=
0
{\displaystyle y^{2}-2(5)\cos(x)y+(5)^{2}-(13)^{2}=0}
y
2
−
10
cos
(
x
)
y
−
144
=
0
{\displaystyle y^{2}-10\cos(x)y-144=0}
y
=
10
cos
(
x
)
+
(
10
cos
(
x
)
)
2
−
4
(
−
144
)
2
{\displaystyle y={\frac {10\cos(x)+{\sqrt {(10\cos(x))^{2}-4(-144)}}}{2}}}
y
=
10
cos
(
x
)
+
100
(
cos
(
x
)
)
2
+
576
2
{\displaystyle y={\frac {10\cos(x)+{\sqrt {100(\cos(x))^{2}+576}}}{2}}}
y
=
5
cos
(
x
)
+
25
(
cos
(
x
)
)
2
+
144
{\displaystyle y=5\cos(x)+{\sqrt {25(\cos(x))^{2}+144}}}
Code supplied to grapher (without white space) is:
( 5 )( cos ( x )) + ( (( 25 )(( cos ( x )) ^ 2 ) + 144 ) ^ ( 0.5 ) )
When expressed in this way, it's easy to convert the code to python code:
( 5 ) * ( cos ( x )) + ( (( 25 ) * (( cos ( x )) ** 2 ) + 144 ) ** ( 0.5 ) )
Positions of interest:
Graph and diagram of piston at top dead center.
Graph and diagram of piston at bottom dead center.
Graph and diagram of piston half-way between TDC and BDC.
Graph and diagram of crank half-way between TDC and BDC.
Graph and diagram of connecting rod tangential to crank.
Figure 4a: Diagram showing acceleration of piston in reciprocating engine. Acceleration introduces the second derivative. While velocity was the first derivative of position with respect to time,
acceleration is the first derivative of velocity or the second derivative of position.
From velocity above
y
y
′
−
5
cos
(
x
)
y
′
+
5
sin
(
x
)
y
=
0
{\displaystyle yy'-5\cos(x)y'+5\sin(x)y=0}
By implicit differentation:
y
y
″
+
y
′
y
′
−
5
(
cos
(
x
)
y
″
+
y
′
(
−
sin
(
x
)
)
)
+
5
(
sin
(
x
)
y
′
+
y
cos
(
x
)
)
=
0
{\displaystyle yy''+y'y'-5(\cos(x)y''+y'(-\sin(x)))+5(\sin(x)y'+y\cos(x))=0}
y
y
″
+
y
′
y
′
−
5
cos
(
x
)
y
″
+
5
y
′
sin
(
x
)
+
5
sin
(
x
)
y
′
+
5
y
cos
(
x
)
=
0
{\displaystyle yy''+y'y'-5\cos(x)y''+5y'\sin(x)+5\sin(x)y'+5y\cos(x)=0}
y
y
″
−
5
cos
(
x
)
y
″
=
−
(
y
′
y
′
+
5
y
′
sin
(
x
)
+
5
sin
(
x
)
y
′
+
5
y
cos
(
x
)
)
{\displaystyle yy''-5\cos(x)y''=-(y'y'+5y'\sin(x)+5\sin(x)y'+5y\cos(x))}
y
″
(
y
−
5
cos
(
x
)
)
=
−
(
y
′
y
′
+
10
sin
(
x
)
y
′
+
5
y
cos
(
x
)
)
{\displaystyle y''(y-5\cos(x))=-(y'y'+10\sin(x)y'+5y\cos(x))}
y
″
=
y
′
y
′
+
10
sin
(
x
)
y
′
+
5
y
cos
(
x
)
5
cos
(
x
)
−
y
{\displaystyle y''={\frac {y'y'+10\sin(x)y'+5y\cos(x)}{5\cos(x)-y}}}
Substitute for
y
{\displaystyle y}
y
′
{\displaystyle y'}
Figure 4b: Diagram showing "irregularities" in the curve of velocity while velocity is increasing.
y
{\displaystyle y}
"Kinks" in the curve:
It is not obvious by looking at the curve of velocity that there are slight irregularities in the curve when velocity is increasing.
However, the irregularities are obvious in the curve of acceleration.
During one revolution of the crankshaft there is less time allocated for negative acceleration than for positive acceleration.
Therefore, the maximum absolute value of negative acceleration is greater than the maximum value of positive acceleration.
Minimum and maximum velocity [ edit | edit source ]
Figure 5: Diagram showing positions of maximum velocity of piston in reciprocating engine.
x
=
0
∘
{\displaystyle x=0^{\circ }}
x
=
90
∘
,
{\displaystyle x=90^{\circ },}
6
{\displaystyle 6}
x
=
90
∘
{\displaystyle x=90^{\circ }}
x
=
180
∘
,
{\displaystyle x=180^{\circ },}
4
{\displaystyle 4}
Velocity is rate of change of position. See also Figure 3 above.
Velocity is zero when slope of curve of position is zero.
This occurs at top dead center and at bottom dead center, ie, when
x
=
0
{\displaystyle x=0}
x
=
π
.
{\displaystyle x=\pi .}
Intuition suggests that the position of maximum velocity might be the point at which the connecting rod is tangent
to the circle of the crankshaft. In other words:
x
=
arctan
(
13
5
)
=
68.96
…
∘
{\displaystyle x=\arctan({\frac {13}{5}})=68.96\dots ^{\circ }}
or, that the position of maximum velocity might be the point at which the piston is half-way between top dead center and bottom dead center.
In other words:
x
=
arccos
(
2.5
13
)
=
78.9
…
∘
{\displaystyle x=\arccos({\frac {2.5}{13}})=78.9\dots ^{\circ }}
0
,
{\displaystyle 0,}
x
=
±
71.26
…
∘
.
{\displaystyle x=\pm 71.26\dots ^{\circ }.}
Suppose that the engine is rotating at
100
{\displaystyle 100}
955
{\displaystyle 955}
v
m
a
x
=
y
′
i
n
c
h
e
s
r
a
d
i
a
n
⋅
100
r
a
d
i
a
n
s
s
e
c
o
n
d
{\displaystyle v_{max}=y'{\frac {inches}{radian}}\cdot 100{\frac {radians}{second}}}
abs
(
v
m
a
x
)
=
5.36
(
100
)
{\displaystyle (v_{max})=5.36(100)}
30.5
{\displaystyle 30.5}
Minimum and maximum acceleration [ edit | edit source ]
Figure 6: Diagram showing positions of minimum and maximum acceleration of piston in reciprocating engine. Acceleration is rate of change of velocity. See also Figures 4a and 4b above.
Acceleration is zero when slope of curve of velocity is zero. This occurs at maximum velocity or when
∠
x
{\displaystyle \angle x}
±
71.3
∘
.
{\displaystyle \pm 71.3^{\circ }.}
Acceleration is maximum when slope of curve of velocity is maximum.
Maximum negative acceleration occurs when slope of curve of velocity is maximum negative. This happens at top dead center
when
∠
x
=
0.
{\displaystyle \angle x=0.}
Maximum positive acceleration occurs when slope of curve of velocity is maximum positive. This happens before and after bottom
dead center when
∠
x
{\displaystyle \angle x}
±
127
∘
.
{\displaystyle \pm 127^{\circ }.}
100
{\displaystyle 100}
a
c
c
m
a
x
=
y
″
i
n
c
h
e
s
r
a
d
i
a
n
2
⋅
100
r
a
d
i
a
n
s
s
e
c
o
n
d
⋅
100
r
a
d
i
a
n
s
s
e
c
o
n
d
{\displaystyle acc_{max}=y''{\frac {inches}{radian^{2}}}\cdot 100{\frac {radians}{second}}\cdot 100{\frac {radians}{second}}}
abs
(
a
c
c
m
a
x
)
=
6.92
∗
100
∗
100
{\displaystyle (acc_{max})=6.92*100*100}
2
{\displaystyle ^{2}}
180
{\displaystyle 180}
This maximum value of acceleration is maximum negative when
∠
x
=
0.
{\displaystyle \angle x=0.}
f
=
m
a
{\displaystyle f=ma}
w
=
f
s
{\displaystyle w=fs}
2
{\displaystyle ^{2}}
Perhaps this helps to explain why a big marine diesel engine rotating at low RPM can achieve efficiency of
55
%
.
{\displaystyle 55\%.}
Let a body move in accordance with the following function of
t
,
f
(
t
)
{\displaystyle t,\ f(t)}
t
{\displaystyle t}
p
t
=
f
(
t
)
=
a
t
2
+
b
t
+
c
{\displaystyle p_{t}=f(t)=at^{2}+bt+c}
p
t
{\displaystyle p_{t}}
t
.
{\displaystyle t.}
p
t
{\displaystyle p_{t}}
f
(
t
)
{\displaystyle f(t)}
For
b
t
{\displaystyle bt}
b
{\displaystyle b}
l
e
n
g
t
h
t
i
m
e
{\displaystyle {\frac {length}{time}}}
For
a
t
2
{\displaystyle at^{2}}
a
{\displaystyle a}
l
e
n
g
t
h
t
i
m
e
2
{\displaystyle {\frac {length}{time^{2}}}}
If
t
==
0
,
p
0
=
c
{\displaystyle t==0,\ p_{0}=c}
p
t
=
f
(
t
)
=
a
t
2
+
b
t
+
p
0
.
{\displaystyle p_{t}=f(t)=at^{2}+bt+p_{0}.}
The derivatives enable us to assign specific values to
a
,
b
.
{\displaystyle a,\ b.}
d
(
p
t
)
d
t
=
f
′
(
t
)
=
v
t
=
2
a
t
+
b
{\displaystyle {\frac {d(p_{t})}{dt}}=f'(t)=v_{t}=2at+b}
v
t
{\displaystyle v_{t}}
t
.
{\displaystyle t.}
If
t
==
0
,
v
0
=
b
{\displaystyle t==0,\ v_{0}=b}
p
t
=
f
(
t
)
=
a
t
2
+
v
0
t
+
p
0
.
{\displaystyle p_{t}=f(t)=at^{2}+v_{0}t+p_{0}.}
d
(
v
t
)
d
t
=
f
″
(
t
)
=
2
a
,
{\displaystyle {\frac {d(v_{t})}{dt}}=f''(t)=2a,}
For convenience let us say that
p
t
=
1
2
a
t
2
+
v
0
t
+
p
0
{\displaystyle p_{t}={\frac {1}{2}}at^{2}+v_{0}t+p_{0}}
a
{\displaystyle a}
Then
v
t
=
1
2
⋅
2
a
t
+
v
0
=
a
t
+
v
0
{\displaystyle v_{t}={\frac {1}{2}}\cdot 2at+v_{0}=at+v_{0}}
d
(
v
t
)
d
t
=
a
.
{\displaystyle {\frac {d(v_{t})}{dt}}=a.}
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