nyuuzyou/wikis huggingface dataset:
Partial Derivative/Examples

== Examples of Partial Derivatives ==
=== Notation for 3-Value Function ===
=== Arbitrary Cubic ===
=== Example: $V = \pi r^2 h$ ===
=== Example: $x z^y$ ===
=== Example: $x^{x y}$ ===
=== Example: $x \map \sin {y z}$ ===
=== Example: $u^2 + x^2 + y^2 = a^2$ ===
=== Example: $u + \ln u = x y$ ===
=== Example: $v + \ln u = x y$, $u + \ln v = x - y$ ===
=== Example: $u - v + 2 w = x + 2 z$, $2 u + v + 2 w = 2 x - 2 z$, $u - v + w = z = y$ ===
=== Example: $2 u + 3 v = \sin x$, $u + 2 v = x \cos y$ ===
=== Example: $u^2 + v^2 = x^2$, $2 u v = 2 x y + y^2$ ===
Category:Examples of Partial Derivatives
Partial Derivative/Examples/2 u + 3 v = sin x, u + 2 v = x cos y

== Example of Partial Derivative ==
Consider the simultaneous equations:
:$\begin {cases} 2 u + 3 v & = \sin x \\ u + 2 v & = x \cos y \end {cases}$
Then:
:$\map {u_1} {\dfrac \pi 2, \pi} = 3$
== Proof ==
By definition of partial derivative:
:$\map {u_1} {\dfrac \pi 2, \pi} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = \frac \pi 2, y \mathop = \pi}$
hence the motivation for the abbreviated notation on the .
=== Explicit Method ===
=== Implicit Method ===
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $14$
Category:Examples of Partial Derivatives
Category:Partial Derivative/Examples/2 u + 3 v = sin x, u + 2 v = x cos y
Partial Derivative/Examples/2 u + 3 v = sin x, u + 2 v = x cos y/Explicit Method

== Example of Partial Derivative ==
== Proof ==
By definition of partial derivative:
:$\map {u_1} {\dfrac \pi 2, \pi} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = \frac \pi 2, y \mathop = \pi}$
hence the motivation for the abbreviated notation on the .
We have:
:$2 u + 3 v = \sin x$
Thus:
Then we have:
:$u + 2 v = x \cos y$
Thus:
Substituting $\dfrac {\partial v} {\partial x}$ from $(2)$ into $(1)$ gives:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $14$
Category:Partial Derivative/Examples/2 u + 3 v = sin x, u + 2 v = x cos y
Partial Derivative/Examples/2 u + 3 v = sin x, u + 2 v = x cos y/Implicit Method

== Example of Partial Derivative ==
== Proof ==
By definition of partial derivative:
:$\map {u_1} {\dfrac \pi 2, \pi} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = \frac \pi 2, y \mathop = \pi}$
hence the motivation for the abbreviated notation on the .
Combining $(1)$ and $(2)$ into matrix form:
:$\begin {pmatrix} 2 & 3 \\ 1 & 2 \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial x} \\ \dfrac {\partial v} {\partial x} \end {pmatrix} = \begin {pmatrix} \cos x \\ \cos y \end {pmatrix}$
Hence by Cramer's Rule:
Hence:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $14$
Category:Partial Derivative/Examples/2 u + 3 v = sin x, u + 2 v = x cos y
Partial Derivative/Examples/Arbitrary Cubic

== Examples of Partial Derivatives ==
Let $\map z {x, y}$ be the real function of $2$ variables defined as:
:$z = x^3 - 3 x y + 2 y^2$
Then we have:
== Proof ==
All results follow from Power Rule for Derivatives and the definition of partial derivative.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
Category:Examples of Partial Derivatives
Partial Derivative/Examples/Notation for 3-Value Function

== Examples of Notation for Partial Derivatives ==
Let $u = \map f {x, y, z}$ be a real function of $3$ variables.
Then the partial derivatives may be expressed variously as:
:$\dfrac {\partial u} {\partial x} = \map {f_1} {x, y, z} = \dfrac {\partial f} {\partial x} = \map {\dfrac \partial {\partial x} f} {x, y, z}$
:$\dfrac {\partial u} {\partial y} = \map {f_2} {x, y, z} = \dfrac {\partial f} {\partial y} = \map {\dfrac \partial {\partial y} f} {x, y, z}$
:$\dfrac {\partial u} {\partial z} = \map {f_3} {x, y, z} = \dfrac {\partial f} {\partial z} = \map {\dfrac \partial {\partial z} f} {x, y, z}$
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives
Category:Examples of Partial Derivatives
Partial Derivative/Examples/Volume of Cylinder

== Example of Partial Derivative ==
Let $V = \pi r^2 h$ be an expression representing the volume of a cylinder whose height is $h$ and whose base has radius $r$.
Then:
:the partial derivative of $V$ $r$ may be expressed as:
::$\dfrac {\partial V} {\partial r} = 2 \pi r h$
:the partial derivative of $V$ $h$ may be expressed as:
::$\dfrac {\partial V} {\partial h} = \pi r^2$
== Sources ==
* 
* 
Category:Examples of Partial Derivatives
Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2

== Example of Partial Derivative ==
Consider the simultaneous equations:
Then:
:$\map {u_1} {1, -2} = 1$
at $u = 1$, $v = 0$.
== Proof ==
By definition of partial derivative:
:$\map {u_1} {1, -2} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = 1, y \mathop = -2}$
hence the motivation for the abbreviated notation on the .
=== Lemma ===
=== Explicit Method ===
=== Implicit Method ===
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $15$
Category:Examples of Partial Derivatives
Category:Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2
Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2/Explicit Method

== Example of Partial Derivative ==
== Proof ==
By definition of partial derivative:
:$\map {u_1} {1, -2} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = 1, y \mathop = -2}$
hence the motivation for the abbreviated notation on the .
=== Lemma ===
We have:
At this point we investigate the sign of $\sqrt {x^2 - v^2}$ which is needed.
We see that:
Hence when we are at the point of plugging in numbers we will need to take the positive square root.
Then we have:
Substituting $\dfrac {\partial v} {\partial x}$ from $(2)$ into $(1)$ gives:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $15$
Category:Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2
Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2/Implicit Method

== Example of Partial Derivative ==
== Proof ==
By definition of partial derivative:
:$\map {u_1} {1, -2} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = 1, y \mathop = -2}$
hence the motivation for the abbreviated notation on the .
=== Lemma ===
We have:
Then we have:
Combining $(1)$ and $(2)$ into matrix form:
:$\begin {pmatrix} u & v \\ v & u \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial x} \\ \dfrac {\partial v} {\partial x} \end {pmatrix} = \begin {pmatrix} x \\ y \end {pmatrix}$
Hence by Cramer's Rule:
Hence:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $15$
Category:Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2
Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2/Lemma

== Lemma for Example: $u^2 + v^2 = x^2$, $2 u v = 2 x y + y^2$ ==
Consider the simultaneous equations:
Then:
:$x = 1$, $y = -2$ is a solution at $u = 1$, $v = 0$.
== Proof ==
We make sure that $\tuple {1, -2}$ is actually a solution at $u = 1$, $v = 0$:
and it is seen that this is the case.
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $15$
Category:Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2
Partial Derivative/Examples/u^2 + x^2 + y^2 = a^2

== Example of Partial Derivative ==
Let $u^2 + x^2 + y^2 = a^2$ be an implicit function.
Then:
== Proof ==
and:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.2$ Implicit Functions
Category:Examples of Partial Derivatives
Partial Derivative/Examples/u - v + 2 w, 2 u + v + 2 w, u - v + w

== Theorem ==
Let:
Then:
== Proof ==
Partial differentiation $y$ gives:
which can be expressed in matrix form as:
:$\begin {pmatrix} 1 & -1 & 2 \\ 2 & 1 & -2 \\ 1 & -1 & 1 \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial y} \\ \dfrac {\partial v} {\partial y} \\ \dfrac {\partial w} {\partial y} \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ -1 \end {pmatrix}$
Solving by Cramer's Rule:
The solution can be read directly:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $5$
Category:Examples of Partial Derivatives
Partial Derivative/Examples/v + ln u = x y, u + ln v = x - y

== Example of Partial Derivative ==
Consider the simultaneous equations:
:$\begin {cases} v + \ln u = x y \\ u + \ln v = x - y \end {cases}$
Then:
=== Second Partial Derivative ===
== Proof ==
and so combining $(1)$ and $(2)$ into matrix form:
:$\begin {pmatrix} 1 & u \\ v & 1 \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial x} \\ \dfrac {\partial v} {\partial x} \end {pmatrix} = \begin {pmatrix} y u \\ v \end {pmatrix}$
Hence by Cramer's Rule:
and:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.2$ Implicit Functions: Example $\text C$
Category:Examples of Partial Derivatives
Partial Derivative/Examples/v + ln u = x y, u + ln v = x - y/Second Partial Derivative

== Example of Partial Derivative ==
Consider the simultaneous equations:
:$\begin {cases} v + \ln u = x y \\ u + \ln v = x - y \end {cases}$
Then:
:$\dfrac {\partial^2 u} {\partial x^2} = \dfrac {u \paren {\paren {y - v}^2 - v \paren {\paren {1 - y u} + u \dfrac {y - v} {1 - u v} \paren {1 + y - v - y u} } } } {\paren {1 - u v}^2}$
== Proof ==
We have from Partial Derivatives of $v + \ln u = x y$, $u + \ln v = x - y$ that:
Hence:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $13$
Category:Examples of Partial Derivatives
Partial Derivative/Examples/x^(x y)

== Example of Partial Derivative ==
Let $\map f {x, y} = x^{x y}$ be a real function of $2$ variables such that $x, y \in \R_{>0}$.
Then:
== Proof ==
=== With Respect to $x$ ===
=== With Respect to $y$ ===
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives: Example $\text A$
Category:Examples of Partial Derivatives
Partial Derivative/Examples/x^(x y)/wrt x

== Example of Partial Derivative ==
Let $\map f {x, y} = x^{x y}$ be a real function of $2$ variables such that $x, y \in \R_{>0}$.
Then:
:$\dfrac {\partial f} {\partial x} = x^{x y} \paren {y \ln x + y}$
== Proof ==
By definition, the partial derivative $x$ is obtained by holding $y$ constant.
Hence Derivative of $x^{a x}$ can be directly used:
:$\dfrac \d {\d x} x^{y x} = y x^{y x} \paren {\ln x + 1}$
The result can then be rearranged to match the form given.
 
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives: Example $\text A$
Category:Examples of Partial Derivatives
Partial Derivative/Examples/x^(x y)/wrt y

== Example of Partial Derivative ==
Let $\map f {x, y} = x^{x y}$ be a real function of $2$ variables such that $x, y \in \R_{>0}$.
Then:
:$\dfrac {\partial f} {\partial y} = x^{x y + 1} \ln x$
== Proof ==
By definition, the partial derivative $y$ is obtained by holding $x$ constant.
From Derivative of Power of Constant:
:$\map {D_y} {x^y} = x^y \ln x$
for constant $a$.
Then:
 
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives: Example $\text A$
Category:Examples of Partial Derivatives
Partial Derivative/Examples/x sine y z

== Example of Partial Derivative ==
Let $\map f {x, y, z} = x \map \sin {y z}$ be a real function of $3$ variables.
Then:
:$\map {f_3} {a, 1, \pi} = -a$
== Proof ==
By definition, the partial derivative the $3$rd variable $z$ is obtained by holding $1$st and $2$nd ones constant.
 
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives: Example $\text B$
Category:Examples of Partial Derivatives
Partial Derivative/Examples/x z^y

== Example of Partial Derivative ==
Let $\map f {x, y, z} = x z^y$ be a real function of $3$ variables.
Then the partial derivative the $2$nd variable may be expressed as:
:$\map {f_2} {x, y, z} = x z^y \ln z$
and because of the notation chosen, we have:
:$\map {f_2} {r, s, t} = r t^s \ln t$
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: $1.1$ Partial Derivatives
Category:Examples of Partial Derivatives
Partial Derivative of Hamiltonian with respect to Displacement

== Theorem ==
Let $H$ be a Hamiltonian of a system:
:$\ds H = -F + \sum_{i \mathop = 1}^n y_i' F_{y_i'}$
where:
:$F$ is the Lagrangian of the system
:$y_i$ are the generalized coordinates
:$y_i'$ is the first derivative of $q_i$ time
:$p_i$ is the momentum of the system in those generalized coordinates:
::$p_i = \dfrac {\partial F} {\partial q_i}$
Then the partial derivative of $H$ momentum is given by:
:$\dfrac {\partial H} {\partial y_i} = -p_i'$
== Proof ==
== Sources ==
* 
* 
Category:Hamiltonians
Category:Examples of Partial Derivatives
Partial Derivative of Hamiltonian with respect to Momentum

== Theorem ==
Let $H$ be a Hamiltonian of a system:
:$\ds H = -F + \sum_{i \mathop = 1}^n y_i' F_{y_i'}$
where:
:$F$ is the Lagrangian of the system
:$y_i$ are the generalized coordinates
:$y_i'$ is the first derivative of $q_i$ time
:$p_i$ is the momentum of the system in those generalized coordinates:
::$p_i = \dfrac {\partial F} {\partial q_i}$
Then the partial derivative of $H$ momentum is given by:
:$\dfrac {\partial H} {\partial p_i} = y_i'$
== Proof ==
== Sources ==
* 
* 
Category:Hamiltonians
Category:Examples of Partial Derivatives
Partial Derivative of Hamiltonian with respect to Time

== Theorem ==
Let $H$ be a Hamiltonian of a system:
:$\ds H = -F + \sum_{i \mathop = 1}^n y_i' F_{y_i'}$
where:
:$F$ is the Lagrangian of the system
:$y_i$ are the generalized coordinates
:$y_i'$ is the first derivative of $q_i$ time
:$p_i$ is the momentum of the system in those generalized coordinates:
::$p_i = \dfrac {\partial F} {\partial q_i}$
Then the partial derivative of $H$ time is given by:
:$\dfrac {\partial H} {\partial t} = \dfrac {\partial F} {\partial t}$
== Proof ==
== Sources ==
* 
* 
Category:Hamiltonians
Category:Examples of Partial Derivatives
Partial Derivative wrt x of sin x y over cos (x + y)

== Theorem ==
:$\dfrac \partial {\partial x} \dfrac {\sin x y} {\map \cos {x + y} } = \dfrac {y \map \cos {x + y} \cos x y + \map \sin {x + y} \sin x y} {\map {\cos^2} {x + y} }$
== Proof ==
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $1$
Category:Examples of Partial Derivatives
Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)

== Theorem ==
Let $z^2 = x^2 - 2 x y - 1$.
Then:
:$\valueat {\dfrac {\partial z} {\partial x} } {x \mathop = 1, y \mathop = -2, z \mathop = -2} = -\dfrac 3 2$
== Proof ==
First we make sure that $\tuple {1, -2, -2}$ actually satisfies the equation:
and it is seen that this is the case.
=== Explicit Method ===
=== Implicit Method ===
Category:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)
Category:Examples of Partial Derivatives
Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Explicit Method

== Theorem ==
== Proof ==
First we make sure that $\tuple {1, -2, -2}$ actually satisfies the equation:
and it is seen that this is the case.
When $z = -2$ we have that $\sqrt {x^2 - 2 x y - 1} < 0$.
Hence:
:$z = -\sqrt {x^2 - 2 x y - 1}$
Thus we continue:
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $4$
Category:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)
Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Implicit Method

== Theorem ==
== Proof ==
First we make sure that $\tuple {1, -2, -2}$ actually satisfies the equation:
and it is seen that this is the case.
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $4$
Category:Examples of Partial Derivatives
Partial Derivatives of Metric vanish at Origin of Normal Neighborhood

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian or pseudo-Riemannian manifold.
Let $U_p$ be the normal neighborhood for $p \in M$.
Let $\struct {U_p, \tuple {x^i}}$ be a normal coordinate chart.
Suppose $g_{ij}$ are the components of metric $g$ in coordinates $\tuple {x^i}$ at $p \in M$.
Then all partial derivatives of $g_{ij}$ vanish at $p$:
:$\map {\dfrac {\partial g_{ij} }{\partial x^k} } {\map {x^r} p} = 0$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Normal Neighborhoods and Normal Coordinates
Category:Normal Neighborhoods
Category:Riemannian Metrics
Category:Pseudo-Riemannian Metrics
Partial Derivatives of Solution of Hamilton-Jacobi Equation are First Integrals of Euler's Equations

== Theorem ==
Let $\mathbf y = \sequence {y_i}_{1 \mathop \le i \mathop \le n}$, $\bsalpha = \sequence {\alpha_i}_{1 \mathop \le i \mathop \le m}$ be vectors, where $m \le n$.
Let $S = \map S {x, \mathbf y, \bsalpha}$ be a solution of the Hamilton-Jacobi equation, where $\bsalpha$ are parameters.
Then each partial derivative:
:$\dfrac {\partial S} {\partial \alpha_i}$
is a first integral of canonical Euler's equations.
== Proof ==
Consider the total derivative of $\dfrac {\partial S} {\partial \alpha_i}$ $x$:
If Euler's equations are satisfied, the vanishes.
Hence
:$\dfrac \d {\d x} \dfrac {\partial S} {\partial \alpha_i} = 0$
or:
:$\dfrac {\partial S} {\partial \alpha_i} = C_i$
where $C_i$ is a constant.
== Sources ==
* : $\S 4.23$: The Hamilton-Jacobi Equation. Jacobi's Theorem 
Category:Calculus of Variations
Partial Derivatives of tan^2 (x^2 - y^2)

== Theorem ==
Let:
:$\map f {x, y} = \map {\tan^2} {x^2 - y^2}$
Then:
== Proof ==
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $3$
Category:Examples of Partial Derivatives
Partial Difference of Integer Combinations is Integer Combination

== Lemma ==
Let $a, b \in \Z$ be integers.
Let $S = \set {a x + b y: x, y \in \Z}$ be the set of integer combinations of $a$ and $b$.
Let $u \in S$ and $v \in S$.
Then:
:$u \mathop {\dot -} v \in S$
where $\dot -$ denotes the extension of the partial subtraction operator to the integers.
== Proof ==
As both $u, v \in S$, $u$ and $v$ can be expressed as:
where $x_1, x_2, y_1, y_2$ are integers.
Let $u \ge v$.
Then:
As Integer Subtraction is Closed, both $x_1 - x_2$ and $y_1 - y_2$ are integers.
Thus $u \mathop {\dot -} v \in S$.
Let $u < v$.
Then by definition of the partial subtraction operator:
:$u \mathop {\dot -} v = 0$
From Set of Integer Combinations includes Zero, $0 \in S$.
Hence the result.
== Sources ==
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Lemma $1$
Category:Integer Combinations
Partial Differential Equation of Planes in 3-Space

== Theorem ==
The set of planes in real Cartesian $3$-dimensional space can be described by the system of partial differential equations:
== Proof ==
From Equation of Plane, we have that the equation defining a general plane $P$ is:
:$\alpha_1 x + \alpha_2 y + \alpha_3 z = \gamma$
which can be written as:
:$z = a x + b y + c$
by setting:
We use the technique of Elimination of Constants by Partial Differentiation.
We see we have:
:$1$ dependent variable, that is: $z$
:$2$ independent variables, that is: $x$ and $y$
:$3$ constants, that is: $a$, $b$ and $c$.
Taking the partial first derivatives $x$ and $y$, we get:
$2$ equations are insufficient to dispose of $3$ constants, so the process continues by taking the partial second derivatives $x$ and $y$:
and the system of partial differential equations has been established.
== Also defined as ==
Some older sources suggest that it is "customary" to assign a standard system of labels to these partial differential equations:
but this is a technique which is rarely emphasised in more modern works.
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.211$ The Partial Differential Equations of All Planes and All Spheres
Category:Partial Differentiation
Category:Solid Analytic Geometry
Partial Differential Equation of Spheres in 3-Space

== Theorem ==
The set of spheres in real Cartesian $3$-dimensional space can be described by the system of partial differential equations:
:$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$
and if the radii of these spheres are expected to be real:
:$z_{xx} z_{yy} > z_{xy}$
== Proof ==
From Equation of Sphere, we have that the equation defining a general sphere $S$ is:
:$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$
where $a$, $b$ and $c$ are arbitrary constants.
We use the technique of Elimination of Constants by Partial Differentiation.
Taking the partial first derivatives $x$ and $y$ and simplifying, we get:
$2$ equations are insufficient to dispose of $3$ constants, so the process continues by taking the partial second derivatives $x$ and $y$:
Eliminating $z - c$:
:$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$
This is our system of partial differential equations describing the set of spheres in real Cartesian $3$-dimensional space.
We now investigate the conditions under which the radii of these spheres are real.
Let $\lambda = \dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_y} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$.
Then:
and so:
:$z_{xx} z_{yy} > z_{xy}$
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.211$ The Partial Differential Equations of All Planes and All Spheres
Category:Partial Differentiation
Category:Solid Analytic Geometry
Partial Fractions Expansion/Examples

== Examples of Partial Fractions Expansions ==
=== Example: $\frac 5 6$ ===
=== Arbitrary Example $1$: $\dfrac 1 {x \paren {x + 1}^2}$ ===
=== Arbitrary Example $2$: $\dfrac {x + 3} {x^2 + 3 x + 2}$ ===
Category:Examples of Partial Fractions Expansions
Partial Fractions Expansion/Examples/5 over 6

== Example of Partial Fractions Expansion ==
$\dfrac 5 6$ has the partial fractions expansion given by:
:$\dfrac 5 6 = \dfrac 1 2 + \dfrac 1 3$
Hence $\dfrac 1 2$ and $\dfrac 1 3$ are partial fractions of $\dfrac 5 6$.
== Sources ==
* 
* 
Category:Examples of Partial Fractions Expansions
Partial Fractions Expansion/Examples/Arbitrary Example 1

== Example of Partial Fractions Expansion ==
:$\dfrac 1 {x \paren {x + 1}^2} = \dfrac 1 x - \dfrac 1 {\paren {x + 1} } - \dfrac 1 {\paren {x + 1}^2}$
== Proof ==
Therefore:
:$\dfrac 1 {x \paren {x + 1}^2} = \dfrac 1 x - \dfrac 1 {\paren {x + 1} } - \dfrac 1 {\paren {x + 1}^2}$
Category:Examples of Partial Fractions Expansions
Partial Fractions Expansion/Examples/Arbitrary Example 2

== Example of Partial Fractions Expansion ==
:$\dfrac {x + 3} {x^2 + 3 x + 2} = \dfrac 2 {x + 1} -\dfrac 1 {x + 2}$
== Proof ==
Therefore:
:$\dfrac {x + 3} {x^2 + 3 x + 2} = \dfrac 2 {x + 1} -\dfrac 1 {x + 2}$
== Sources ==
* 
* 
Category:Examples of Partial Fractions Expansions
Partial Fractions Expansion of Cotangent

== Theorem ==
Let $x \in \R \setminus \Z$, that is such that $x$ is a real number that is not an integer.
Then:
:$\ds \pi \cot \pi x = \dfrac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Cotangent Function
Category:Partial Fractions Expansion of Cotangent
Partial Fractions Expansion of Cotangent/Proof 1

== Theorem ==
== Proof ==
We have that:
:$\cot \pi x = \dfrac {\cos \pi x} {\sin \pi x}$
has a denominator which is $0$ at $x = 0, \pm 1, \pm 2, \ldots$.
Hence the limitation on the domain of $x \cot \pi x$ to exclude integer $x$.
Having established that, we should be able to express $\cot \pi x$ in the form:
:$\cot \pi x = \dfrac a x + \ds \sum_{n \mathop = 1}^\infty \paren {\frac {b_n} {x - n} + \frac {c_n} {x + n} }$
using a partial fractions expansion.
By evaluating the coefficients $b_n$ and $c_n$ in the usual manner, they are found to be:
:$\forall n \in \N: b_n = c_n = \dfrac 1 \pi$
The result follows.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.20$: The Bernoulli Numbers and some Wonderful Discoveries of Euler
Category:Partial Fractions Expansion of Cotangent
Partial Fractions Expansion of Cotangent/Proof 1/Historical Note

== Historical Note on Partial Fractions Expansion of Cotangent/Proof 1 ==
The original demonstration of the partial fractions expansion of the cotangent function was presented by .
It is non-rigorous, in the sense that it has not at this stage been established that it is in fact possible to perform that expansion of $\cot \pi x$ into such an expansion in the first place.
== Sources ==
* : Chapter $\text {B}.20$: The Bernoulli Numbers and some Wonderful Discoveries of Euler
Category:Historical Notes