d
f
d
x
=
∂
f
∂
x
+
∑
j
=
1
k
∂
y
j
∂
x
∂
f
∂
y
j
,
{\displaystyle {\frac {\mathrm {d} f}{\mathrm {d} x}}={\frac {\partial f}{\partial x}}+\sum _{j=1}^{k}{\frac {\partial y_{j}}{\partial x}}{\frac {\partial f}{\partial y_{j}}},}
where
f
{\displaystyle f}
is a function of
k
{\displaystyle k}
variables.
d
f
d
x
=
∂
f
∂
x
+
∂
y
∂
x
∂
f
∂
y
,
{\displaystyle {\frac {\mathrm {d} f}{\mathrm {d} x}}={\frac {\partial f}{\partial x}}+{\frac {\partial y}{\partial x}}{\frac {\partial f}{\partial y}},}
where
f
:
R
,
R
→
R
{\displaystyle f:\mathbb {R} ,\mathbb {R} \to \mathbb {R} }
.
Given
f
(
x
,
y
)
=
x
3
−
y
2
,
{\displaystyle f(x,y)=x^{3}-y^{2}\,,}
find
d
f
d
x
.
{\displaystyle \textstyle {\frac {\mathrm {d} f}{\mathrm {d} x}}.}
Begin with the definition of the total derivative:
d
f
d
x
=
∂
f
∂
x
+
∂
y
∂
x
∂
f
∂
y
{\displaystyle {\textstyle {\frac {\mathrm {d} f}{\mathrm {d} x}}={\frac {\partial f}{\partial x}}+{\frac {\partial y}{\partial x}}{\frac {\partial f}{\partial y}}}}
. Notice that in order to continue, we need to calculate
∂
f
∂
x
,
∂
y
∂
x
,
{\displaystyle \textstyle {\frac {\partial f}{\partial x}},\,\textstyle {\frac {\partial y}{\partial x}},\,}
and
∂
f
∂
y
.
{\displaystyle \textstyle {\frac {\partial f}{\partial y}}.}
∂
f
∂
x
=
∂
∂
x
(
x
3
)
=
3
x
2
{\displaystyle {\begin{aligned}{\frac {\partial f}{\partial x}}&={\frac {\partial }{\partial x}}\left(x^{3}\right)\\&=3x^{2}\end{aligned}}}
∂
f
∂
y
=
∂
∂
x
(
y
2
)
=
−
2
y
{\displaystyle {\begin{aligned}{\frac {\partial f}{\partial y}}&={\frac {\partial }{\partial x}}\left(y^{2}\right)\\&=-2y\end{aligned}}}
y
2
=
x
3
2
y
∂
y
∂
x
=
3
x
2
∂
y
∂
x
=
3
x
2
2
y
{\displaystyle {\begin{aligned}y^{2}&=x^{3}\\2y{\frac {\partial y}{\partial x}}&=3x^{2}\\{\frac {\partial y}{\partial x}}&={\frac {3x^{2}}{2y}}\end{aligned}}}
Plugging the results into the definition,
d
f
d
x
=
3
x
2
+
3
x
2
−
2
y
(
2
y
)
{\displaystyle {\textstyle {\frac {\mathrm {d} f}{\mathrm {d} x}}=3x^{2}+{\frac {3x^{2}}{-2y}}\left(2y\right)}}
, we find that
d
f
d
x
=
0
.
{\displaystyle {\textstyle {\frac {\mathrm {d} f}{\mathrm {d} x}}=0}.}
L
=
0
+
1
0
+
1
0
+
1
0
+
1
⋱
{\displaystyle L=0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{\ddots \,}}}}}}}}}
1
L
=
1
0
+
1
0
+
1
0
+
1
0
+
1
⋱
=
L
{\displaystyle {\frac {1}{L}}={\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{\ddots \,}}}}}}}}}}=L}
L
2
=
1
L
=
±
1
{\displaystyle {\begin{aligned}L^{2}&=1\\L&=\pm {\sqrt {1}}\end{aligned}}}
Because
L
{\displaystyle L}
can't be negative,
L
=
1
{\displaystyle L=1}
.
∴
0
+
1
0
+
1
0
+
1
0
+
1
⋱
=
1
{\displaystyle \therefore \ 0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{\ddots \,}}}}}}}}=1}
∀
a
,
b
∈
N
{\displaystyle \forall a,b\in \mathbb {N} }
a
⋅
b
=
a
+
a
⋅
(
b
−
1
)
,
a
⋅
0
=
0
a
b
=
a
⋅
a
b
−
1
,
a
0
=
1
a
↑↑
b
=
a
a
↑↑
(
b
−
1
)
,
a
↑↑
0
=
1
{\displaystyle {\begin{array}{lcll}a\cdot b&=&a+a\cdot (b-1)&,a\cdot 0=0\\a^{b}&=&a\cdot a^{b-1}&,a^{0}=1\\a\uparrow \uparrow b&=&a^{a\uparrow \uparrow (b-1)}&,a\uparrow \uparrow 0=1\end{array}}}
The derivative of a polynomial,
d
d
t
:
P
n
→
P
n
{\displaystyle {\frac {d}{dt}}:\mathbf {P} ^{n}\to \mathbf {P} ^{n}}
,
can be defined as
d
d
t
(
a
0
+
a
1
x
+
a
2
x
2
+
⋯
+
a
n
x
n
)
=
a
1
+
a
2
x
+
a
3
x
2
+
⋯
+
a
n
x
n
−
1
{\displaystyle {\frac {d}{dt}}(a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n})=a_{1}+a_{2}x+a_{3}x^{2}+\cdots +a_{n}x^{n-1}}
.
If we use the standard ordered basis
e
=
{
1
,
x
,
x
2
,
…
,
x
n
}
{\displaystyle \mathbf {e} =\{1,x,x^{2},\ldots ,x^{n}\}}
,
then
(
a
0
+
a
1
x
+
a
2
x
2
+
⋯
+
a
n
x
n
)
e
{\displaystyle (a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n})_{\mathbf {e} }}
can be written as
(
a
0
a
1
a
2
⋮
a
n
)
{\displaystyle {\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}}
,
and
d
d
t
{\displaystyle {\frac {d}{dt}}}
as
d
d
t
:
(
a
0
a
1
a
2
⋮
a
n
)
↦
(
a
1
a
2
⋮
a
n
0
)
{\displaystyle {\frac {d}{dt}}:{\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}\mapsto {\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\\0\end{pmatrix}}}
.
Since
A
=
(
0
1
0
⋯
0
0
0
1
⋯
0
⋮
⋮
⋮
⋱
⋮
0
0
0
⋯
1
0
0
0
⋯
0
)
{\displaystyle A={\begin{pmatrix}0&1&0&\cdots &0\\0&0&1&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\cdots &1\\0&0&0&\cdots &0\\\end{pmatrix}}}
satisfies
A
(
a
0
a
1
a
2
⋮
a
n
)
=
(
a
1
a
2
⋮
a
n
0
)
{\displaystyle A{\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\\0\end{pmatrix}}}
,
A
{\displaystyle A}
represents
d
d
t
{\displaystyle {\frac {d}{dt}}}
.
u
=
u
1
e
1
+
u
2
e
2
+
u
3
e
3
{\displaystyle \mathbf {u} =u_{1}\mathbf {e_{1}} +u_{2}\mathbf {e_{2}} +u_{3}\mathbf {e_{3}} }
v
=
v
1
e
1
+
v
2
e
2
+
v
3
e
3
{\displaystyle \mathbf {v} =v_{1}\mathbf {e_{1}} +v_{2}\mathbf {e_{2}} +v_{3}\mathbf {e_{3}} }
u
∧
v
=
(
u
1
v
2
−
v
1
u
2
)
∧
e
1
+
(
u
2
v
3
−
v
2
u
3
)
∧
e
2
+
(
u
3
v
1
−
v
1
u
3
)
∧
e
3
=
u
1
v
1
{\displaystyle {\begin{aligned}\mathbf {u} \wedge \mathbf {v} &=(u_{1}v_{2}-v_{1}u_{2})\wedge \mathbf {e_{1}} +(u_{2}v_{3}-v_{2}u_{3})\wedge \mathbf {e_{2}} +(u_{3}v_{1}-v_{1}u_{3})\wedge \mathbf {e_{3}} \\&=u_{1}v_{1}\end{aligned}}}
General second degree linear ordinary differential equation [ edit ]
A second degree linear ordinary differential equation is given by
y
″
+
a
(
x
)
y
′
+
b
(
x
)
y
=
c
(
x
)
.
{\displaystyle y''+a(x)y'+b(x)y=c(x).\,}
One way to solve this is to look for some integrating factor ,
M
{\displaystyle M}
, such that
M
y
″
+
M
a
(
x
)
y
′
+
M
b
(
x
)
y
=
(
M
y
)
″
=
M
c
(
x
)
.
{\displaystyle My''+Ma(x)y'+Mb(x)y=(My)''=Mc(x).\,}
Expanding
(
M
y
)
″
{\displaystyle (My)''}
and setting it equal to
M
y
″
+
M
a
(
x
)
y
′
+
M
b
(
x
)
y
=
M
y
″
+
2
M
y
′
+
M
″
y
{\displaystyle My''+Ma(x)y'+Mb(x)y=My''+2My'+M''y\,}
{
2
M
′
=
M
a
(
x
)
M
″
=
M
b
(
x
)
{\displaystyle \left\{{\begin{aligned}2M'&=Ma(x)\\M''&=Mb(x)\\\end{aligned}}\right.}
M
″
a
(
x
)
=
2
M
′
b
(
x
)
{\displaystyle M''a(x)=2M'b(x)}
u
=
M
′
{\displaystyle u=M'}
d
u
d
x
a
(
x
)
=
2
u
b
(
x
)
{\displaystyle {\frac {du}{dx}}a(x)=2ub(x)}
1
2
∫
d
u
u
=
∫
b
(
x
)
a
(
x
)
d
x
{\displaystyle {\frac {1}{2}}{\int {\frac {du}{u}}}={\int {\frac {b(x)}{a(x)}}dx}}
1
2
ln
u
=
∫
b
(
x
)
a
(
x
)
d
x
{\displaystyle {\frac {1}{2}}\ln u={\int {\frac {b(x)}{a(x)}}dx}}
u
=
e
2
∫
b
(
x
)
a
(
x
)
d
x
{\displaystyle u=e^{2{\int {\frac {b(x)}{a(x)}}dx}}}
M
′
=
e
2
∫
b
(
x
)
a
(
x
)
d
x
{\displaystyle M'=e^{2{\int {\frac {b(x)}{a(x)}}dx}}}
M
=
∫
e
2
∫
b
(
x
)
a
(
x
)
d
x
d
x
{\displaystyle M={\int e^{2{\int {\frac {b(x)}{a(x)}}dx}}dx}}
(
M
y
)
″
=
M
c
(
x
)
(
M
y
)
′
=
∫
M
c
(
x
)
d
x
+
C
1
M
y
=
∫
∫
M
c
(
x
)
d
x
d
x
+
C
1
x
+
C
2
{\displaystyle {\begin{aligned}(My)''&=Mc(x)\\(My)'&={\int Mc(x)dx}+C_{1}\\My&={\int {\int Mc(x)dx}dx}+C_{1}x+C_{2}\\\end{aligned}}}
y
=
∫
∫
∫
c
(
x
)
e
2
∫
b
(
x
)
a
(
x
)
d
x
d
x
d
x
d
x
+
C
1
x
+
C
2
∫
e
2
∫
b
(
x
)
a
(
x
)
d
x
d
x
{\displaystyle y={\frac {{\int {\int {\int c(x)e^{2{\int {\frac {b(x)}{a(x)}}dx}}dx}dx}dx}+C_{1}x+C_{2}}{\int e^{2{\int {\frac {b(x)}{a(x)}}dx}}dx}}}
Differential example [ edit ]
The key to differentials is to think of
x
{\displaystyle x}
as a function from some real number
p
{\displaystyle p}
to itself; and
d
x
{\displaystyle dx}
as a function of some that same real number
p
{\displaystyle p}
to a linear map
P
:
R
↦
R
.
{\displaystyle \mathbf {P} :\mathbb {R} \mapsto \mathbb {R} .}
Since all linear maps from
R
{\displaystyle \mathbb {R} }
to
R
{\displaystyle \mathbb {R} }
can be written as a
1
×
1
{\displaystyle 1\times 1}
matrix, we can define
P
{\displaystyle \mathbf {P} }
as
[
p
]
{\displaystyle [p]}
and
d
x
{\displaystyle dx}
as
d
x
:
R
→
R
1
×
1
{\displaystyle dx:\mathbb {R} \rightarrow \mathbb {R} ^{1\times 1}}
d
x
:
p
↦
[
1
]
{\displaystyle dx:p\mapsto [1]}
(As a side note, the value of
d
x
{\displaystyle dx}
, and similarly for all differentials, at
p
{\displaystyle p}
is usually written
d
x
p
{\displaystyle dx_{p}}
.)
Without loss of generality, let's take the function
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
. Differentiating, we have
d
f
d
x
=
2
x
.
{\displaystyle {\frac {df}{dx}}=2x.}
Since we defined
d
x
p
{\displaystyle dx_{p}}
as
[
1
]
{\displaystyle [1]}
and
x
(
p
)
{\displaystyle x(p)}
as
p
{\displaystyle p}
, we can rewrite the derivative as
d
f
p
[
1
]
−
1
=
2
x
(
p
)
.
{\displaystyle df_{p}[1]^{-1}=2x(p).\,}
Multiplying both sides by
[
1
]
{\displaystyle [1]}
, we have
d
f
p
=
2
x
(
p
)
[
1
]
.
{\displaystyle df_{p}=2x(p)[1].\,}
And voilà! We can say that for any function
f
:
R
→
R
{\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }
,
d
f
p
=
[
f
′
(
x
(
p
)
)
]
=
f
′
(
x
(
p
)
)
d
x
p
{\displaystyle df_{p}=\left[f'(x(p))\right]=f'(x(p))dx_{p}}