Total derivative

${\displaystyle {\frac {\mathrm {d} f}{\mathrm {d} x))={\frac {\partial f}{\partial x))+\sum _{j=1}^{k}{\frac {\partial y_{j)){\partial x)){\frac {\partial f}{\partial y_{j))},}$

where ${\displaystyle f}$ is a function of ${\displaystyle k}$ variables.

Example 1

${\displaystyle {\frac {\mathrm {d} f}{\mathrm {d} x))={\frac {\partial f}{\partial x))+{\frac {\partial y}{\partial x)){\frac {\partial f}{\partial y)),}$

where ${\displaystyle f:\mathbb {R} ,\mathbb {R} \to \mathbb {R} }$.

Example 2

Given ${\displaystyle f(x,y)=x^{3}-y^{2}\,,}$ find ${\displaystyle \textstyle {\frac {\mathrm {d} f}{\mathrm {d} x)).}$

Begin with the definition of the total derivative: ${\displaystyle {\textstyle {\frac {\mathrm {d} f}{\mathrm {d} x))={\frac {\partial f}{\partial x))+{\frac {\partial y}{\partial x)){\frac {\partial f}{\partial y))))$. Notice that in order to continue, we need to calculate ${\displaystyle \textstyle {\frac {\partial f}{\partial x)),\,\textstyle {\frac {\partial y}{\partial x)),\,}$ and ${\displaystyle \textstyle {\frac {\partial f}{\partial y)).}$

Plugging the results into the definition, ${\displaystyle {\textstyle {\frac {\mathrm {d} f}{\mathrm {d} x))=3x^{2}+{\frac {3x^{2)){-2y))\left(2y\right)))$, we find that ${\displaystyle {\textstyle {\frac {\mathrm {d} f}{\mathrm {d} x))=0}.}$

Continued fractions

${\displaystyle {\begin{aligned}L^{2}&=1\\L&=\pm {\sqrt {1))\end{aligned))}$

Because ${\displaystyle L}$ can't be negative, ${\displaystyle L=1}$.

${\displaystyle \therefore \ 0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{0+{\cfrac {1}{\ddots \,))))))))=1}$

Tetration and beyond

• ${\displaystyle \forall a,b\in \mathbb {N} }$

  ${\displaystyle {\begin{array}{lcll}a\cdot b&=&a+a\cdot (b-1)&,a\cdot 0=0\\a^{b}&=&a\cdot a^{b-1}&,a^{0}=1\\a\uparrow \uparrow b&=&a^{a\uparrow \uparrow (b-1)}&,a\uparrow \uparrow 0=1\end{array))}$

Polynomials and their derivatives

The derivative of a polynomial,

${\displaystyle {\frac {d}{dt)):\mathbf {P} ^{n}\to \mathbf {P} ^{n))$,

can be defined as

${\displaystyle {\frac {d}{dt))(a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n})=a_{1}+a_{2}x+a_{3}x^{2}+\cdots +a_{n}x^{n-1))$.

If we use the standard ordered basis

${\displaystyle \mathbf {e} =\{1,x,x^{2},\ldots ,x^{n}\))$,

then

${\displaystyle (a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n})_{\mathbf {e} ))$

can be written as

${\displaystyle {\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix))}$,

and ${\displaystyle {\frac {d}{dt))}$ as

${\displaystyle {\frac {d}{dt)):{\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix))\mapsto {\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\\0\end{pmatrix))}$.

Since

${\displaystyle A={\begin{pmatrix}0&1&0&\cdots &0\\0&0&1&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\cdots &1\\0&0&0&\cdots &0\\\end{pmatrix))}$

satisfies

${\displaystyle A{\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix))={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\\0\end{pmatrix))}$

, ${\displaystyle A}$ represents ${\displaystyle {\frac {d}{dt))}$.

Wedge product

${\displaystyle \mathbf {u} =u_{1}\mathbf {e_{1)) +u_{2}\mathbf {e_{2)) +u_{3}\mathbf {e_{3)) }$

${\displaystyle \mathbf {v} =v_{1}\mathbf {e_{1)) +v_{2}\mathbf {e_{2)) +v_{3}\mathbf {e_{3)) }$

${\displaystyle {\begin{aligned}\mathbf {u} \wedge \mathbf {v} &=(u_{1}v_{2}-v_{1}u_{2})\wedge \mathbf {e_{1)) +(u_{2}v_{3}-v_{2}u_{3})\wedge \mathbf {e_{2)) +(u_{3}v_{1}-v_{1}u_{3})\wedge \mathbf {e_{3)) \\&=u_{1}v_{1}\end{aligned))}$

General second degree linear ordinary differential equation

A second degree linear ordinary differential equation is given by

${\displaystyle y''+a(x)y'+b(x)y=c(x).\,}$

One way to solve this is to look for some integrating factor, ${\displaystyle M}$, such that

${\displaystyle My''+Ma(x)y'+Mb(x)y=(My)''=Mc(x).\,}$

Expanding ${\displaystyle (My)''}$ and setting it equal to

${\displaystyle My''+Ma(x)y'+Mb(x)y=My''+2My'+M''y\,}$

${\displaystyle \left\((\begin{aligned}2M'&=Ma(x)\\M''&=Mb(x)\\\end{aligned))\right.}$

${\displaystyle M''a(x)=2M'b(x)}$

${\displaystyle u=M'}$

${\displaystyle {\frac {du}{dx))a(x)=2ub(x)}$

${\displaystyle {\frac {1}{2)){\int {\frac {du}{u))}={\int {\frac {b(x)}{a(x)))dx))$

${\displaystyle {\frac {1}{2))\ln u={\int {\frac {b(x)}{a(x)))dx))$

${\displaystyle u=e^{2{\int {\frac {b(x)}{a(x)))dx))}$

${\displaystyle M'=e^{2{\int {\frac {b(x)}{a(x)))dx))}$

${\displaystyle M={\int e^{2{\int {\frac {b(x)}{a(x)))dx))dx))$

${\displaystyle {\begin{aligned}(My)''&=Mc(x)\\(My)'&={\int Mc(x)dx}+C_{1}\\My&={\int {\int Mc(x)dx}dx}+C_{1}x+C_{2}\\\end{aligned))}$

${\displaystyle y={\frac ((\int {\int {\int c(x)e^{2{\int {\frac {b(x)}{a(x)))dx))dx}dx}dx}+C_{1}x+C_{2)){\int e^{2{\int {\frac {b(x)}{a(x)))dx))dx))}$

Differential example

The key to differentials is to think of ${\displaystyle x}$ as a function from some real number ${\displaystyle p}$ to itself; and ${\displaystyle dx}$ as a function of some that same real number ${\displaystyle p}$ to a linear map ${\displaystyle \mathbf {P} :\mathbb {R} \mapsto \mathbb {R} .}$ Since all linear maps from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$ can be written as a ${\displaystyle 1\times 1}$ matrix, we can define ${\displaystyle \mathbf {P} }$ as ${\displaystyle [p]}$ and ${\displaystyle dx}$ as

${\displaystyle dx:\mathbb {R} \rightarrow \mathbb {R} ^{1\times 1))$

${\displaystyle dx:p\mapsto [1]}$

(As a side note, the value of ${\displaystyle dx}$, and similarly for all differentials, at ${\displaystyle p}$ is usually written ${\displaystyle dx_{p))$.)

Without loss of generality, let's take the function ${\displaystyle f(x)=x^{2))$. Differentiating, we have

${\displaystyle {\frac {df}{dx))=2x.}$

Since we defined ${\displaystyle dx_{p))$ as ${\displaystyle [1]}$ and ${\displaystyle x(p)}$ as ${\displaystyle p}$, we can rewrite the derivative as

${\displaystyle df_{p}[1]^{-1}=2x(p).\,}$

Multiplying both sides by ${\displaystyle [1]}$, we have

${\displaystyle df_{p}=2x(p)[1].\,}$

And voilà! We can say that for any function ${\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }$,

${\displaystyle df_{p}=\left[f'(x(p))\right]=f'(x(p))dx_{p))$