nyuuzyou/wikis huggingface dataset:
Derivative of Sine Function

== Theorem ==
:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Also see ==
* Derivative of Cosine Function
* Derivative of Tangent Function
* Derivative of Cotangent Function
* Derivative of Secant Function
* Derivative of Cosecant Function
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $4$.
* : $\text {II}$. Calculus: Differentiation
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $3.$ Trigonometric functions
* 
* : Appendix $2$: Table of derivatives and integrals of common functions: Trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
Category:Derivatives of Trigonometric Functions
Category:Sine Function
Category:Derivative of Sine Function
Derivative of Sine Function/Corollary

== Corollary to Derivative of Sine Function ==
:$\map {\dfrac \d {\d x} } {\sin a x} = a \cos a x$
== Proof ==
== Also see ==
* Derivative of $\cos a x$
* Derivative of $\tan a x$
* Derivative of $\cot a x$
* Derivative of $\sec a x$
* Derivative of $\csc a x$
Category:Derivative of Sine Function
Derivative of Sine Function/Proof 1

== Theorem ==
== Proof ==
From the definition of the sine function, we have:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
From Radius of Convergence of Power Series over Factorial, this series converges for all $x$.
From Power Series is Differentiable on Interval of Convergence:
The result follows from the definition of the cosine function.
== Sources ==
* : $\S 16.3 \ (1) \ \text{(vi)}$
Category:Derivative of Sine Function
Derivative of Sine Function/Proof 5

== Theorem ==
== Proof ==
Note that we get the same answer as Derivative of Arcsine Function.
By definition of real $\arcsin$ function, $\arcsin$ is bijective on its domain $\closedint {-1} 1$.
Thus its inverse is itself a mapping.
From Inverse of Inverse of Bijection, its inverse is the $\sin$ function.
== Sources ==
* 
Category:Derivative of Sine Function
Derivative of Sine Integral Function

== Theorem ==
:$\dfrac \d {\d x} \paren {\map \Si x} = \dfrac {\sin x} x$
where $\Si$ denotes the sine integral function.
== Proof ==
We have, by the definition of the sine integral function: 
:$\ds \map \Si x = \int_0^x \frac {\sin t} t \rd t$
By Fundamental Theorem of Calculus (First Part): Corollary, we have: 
:$\ds \frac \d {\d x} \paren {\map \Si x} = \frac {\sin x} x$
== Also see ==
* Derivative of Cosine Integral Function
Category:Sine Integral Function
Category:Derivatives
Derivative of Sine of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\sin u} = \cos u \dfrac {\d u} {\d x}$
where $\sin$ is the sine function and $\cos$ is the cosine function.
== Proof ==
== Also see ==
* Derivative of Cosine of Function
* Derivative of Tangent of Function
* Derivative of Cotangent of Function
* Derivative of Secant of Function
* Derivative of Cosecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.14$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $3$
Category:Derivative of Sine Function
Derivative of Solution to Constant Coefficient Homogeneous LSOODE is also Solution

== Theorem ==
Let:
:$(1): \quad y + p y' + q y = 0$
be a constant coefficient homogeneous linear second order ODE.
Let $\map y x$ be a particular solution of $(1)$.
Then its derivative $\map {y'} x$ is also a particular solution of $(1)$.
== Proof ==
By Solution of Constant Coefficient Homogeneous LSOODE, $y$ is in one of the following forms:
:$y = \begin{cases}
C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\
 & \\
C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\
 & \\
e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q
\end{cases}$
Let:
:$y = A e^{m_1 x} + B e^{m_2 x}$
where $A, B \in \R$.
Then:
:$y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$
which is also in the form:
:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
and so is also a particular solution of $(1)$.
Let:
:$y = A e^{m_1 x} + B x e^{m_1 x}$
where $A, B \in \R$.
Then:
which is also in the form:
:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
and so is also a particular solution of $(1)$.
Let:
:$y = e^{a x} \left({A \sin b x + B \cos b x}\right)$
where $A, B \in \R$.
Then:
:$y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$
which is also in the form:
:$y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$
and so is also a particular solution of $(1)$.
Hence the result.
== Sources ==
* : $\S 3.17$: Problem $3$
Category:Constant Coefficient Homogeneous LSOODEs
Derivative of Square Function

== Theorem ==
Let $f: \R \to \R$ be the square function:
:$\forall x \in \R: \map f x = x^2$
Then the derivative of $f$ is given by:
:$\map {f'} x = 2 x$
== Proof 1 ==
== Proof 2 ==
Category:Derivative of Square Function
Category:Derivatives
Category:Square Function
Derivative of Square of Vector-Valued Function

== Theorem ==
Let $\mathbf a: \R \to \R^n$ be a differentiable vector-valued function.
The derivative of its square is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a^2} = 2 \mathbf a \cdot \dfrac {\d \mathbf a} {\d x} = 2 a \dfrac {\d a} {\d x}$
where $a = \norm {\mathbf a}$ is the magnitude of $\mathbf a$.
== Proof ==
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products: $(3.3 a)$
Category:Dot Product
Category:Vector Calculus
Derivative of Subset is Subset of Derivative

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$, $B$ be subsets of $S$.
Then
:$A \subseteq B \implies A' \subseteq B'$
where $A'$ denotes the derivative of $A$ in $T$.
== Proof ==
Let $A \subseteq B$.
Let $x \in A'$.
By Characterization of Derivative by Open Sets it is enough to prove that:
:for every open set $G$ of $T$:
::if $x \in G$
::then there exists $y$ such that $y \in B \cap G$ and $x \ne y$.
Let $G$ be an open set of $T$.
Let $x \in G$.
Then by Characterization of Derivative by Open Sets:
:there exists a point $y$ of $T$ such that $y \in A \cap G$ and $x \ne y$.
By the corollary to Set Intersection Preserves Subsets:
:$A \cap G \subseteq B \cap G$
Hence:
:$y \in B \cap G$ and $x \ne y$.
The conditions of the hypothesis are thus fulfilled, and:
:$x \in B'$
Thus by definition of subset:
:$A' \subseteq B'$
== Sources ==
* 
* 
Category:Set Derivatives
Derivative of Sum of Vector-Valued Functions

== Theorem ==
Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be differentiable vector-valued functions.
Then the derivative of $\map {\mathbf v} t = \map {\mathbf a} t + \map {\mathbf b} t$ is given by:
:$\dfrac {\d \mathbf v} {\d t} = \map {\dfrac \d {\d t} } {\mathbf a + \mathbf b} = \dfrac {\d \mathbf a} {\d t} + \dfrac {\d \mathbf b} {\d t}$
== Proof ==
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products: $(3.2)$
Category:Vector Calculus
Derivative of Tangent Function

== Theorem ==
:$\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x = \dfrac 1 {\cos^2 x}$
when $\cos x \ne 0$.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Derivative of Sine Function
* Derivative of Cosine Function
* Derivative of Cotangent Function
* Derivative of Secant Function
* Derivative of Cosecant Function
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $6$.
* : $\text {II}$. Calculus: Differentiation
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $3.$ Trigonometric functions
* : Appendix $2$: Table of derivatives and integrals of common functions: Trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
Category:Derivative of Tangent Function
Category:Tangent Function
Derivative of Tangent Function/Corollary 1

== Corollary to Derivative of Tangent Function ==
:$\map {\dfrac \d {\d x} } {\tan a x} = a \sec^2 a x$
== Proof ==
== Also see ==
* Derivative of $\sin a x$
* Derivative of $\cos a x$
* Derivative of $\cot a x$
* Derivative of $\sec a x$
* Derivative of $\csc a x$
Category:Derivative of Tangent Function
Derivative of Tangent Function/Corollary 2

== Corollary to Derivative of Tangent Function ==
:$\dfrac \d {\d x} \tan x = 1 + \tan^2 x$
== Proof ==
== Sources ==
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $3.$ Trigonometric functions
Category:Derivative of Tangent Function
Derivative of Tangent Function/Corollary 3

== Corollary to Derivative of Tangent Function ==
:$\map {\dfrac \d {\d x} } {\tan a x} = a \paren {\tan^2 a x + 1}$
== Proof ==
== Also see ==
* Derivative of $\sin a x$
* Derivative of $\cos a x$
* Derivative of $\cot a x$
* Derivative of $\sec a x$
* Derivative of $\csc a x$
Category:Derivative of Tangent Function
Derivative of Tangent Function/Proof 1

== Theorem ==
== Proof == 
From the definition of the tangent function:
:$\tan x = \dfrac {\sin x} {\cos x}$
From Derivative of Sine Function:
:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$
From Derivative of Cosine Function:
:$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
Then:
This is valid only when $\cos x \ne 0$.
The result follows from the Secant is Reciprocal of Cosine.
== Proof ==
* : $\text {II}$. Calculus: Differentiation: Quotient
Category:Derivative of Tangent Function
Derivative of Tangent of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\tan u} = \sec^2 u \dfrac {\d u} {\d x}$
where $\tan$ is the tangent function and $\sec$ is the secant function.
== Proof ==
== Also see ==
* Derivative of Sine of Function
* Derivative of Cosine of Function
* Derivative of Cotangent of Function
* Derivative of Secant of Function
* Derivative of Cosecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.16$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $5$
Category:Derivative of Tangent Function
Derivative of Uniform Limit of Analytic Functions

== Theorem ==
Let $U$ be an open subset of $\C$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of analytic functions $f_n : U \to \C$.
Let $\sequence {f_n}$ converge locally uniformly to $f$ on $U$.
Let $f'$ denote the derivative of $f$.
Then the sequence $\sequence { {f_n}'}_{n \mathop \in \N}$ converges locally uniformly to $f'$.
== Proof ==
Let $a \in U$.
By definition of locally uniform convergence, there exists an open disk $\map {D_{2 r} } a \subseteq U$ such that $f_n$ converges uniformly to $f$ on $\map {D_{2 r} } a$.
That is:
:$\ds (1): \quad \lim_{n \mathop \to \infty} \sup_{z \mathop \in \map {D_{2 r} } a} \cmod {\map {f_n} z - \map f z} = 0$ 
We shall show that:
:$\ds \lim_{n \mathop \to \infty} \sup_{w \mathop \in \map {D_r} a} \cmod {\map {f'_n} w - \map {f'} w} = 0$
Let $w \in \map {D_r} a$.
By Triangle Inequality for Complex Numbers:
:$(2): \quad \map {D_r} w \subseteq \map {D_{2 r} } a \subseteq U$
Thus by Cauchy's Integral Formula for Derivatives:
Therefore:
Since $w \in \map {D_r} a$ was arbitrary, we have:
== Also see ==
* Uniform Limit of Analytic Functions is Analytic
Category:Complex Analysis
Category:Uniform Convergence
Derivative of Uniformly Convergent Sequence of Differentiable Functions

== Theorem ==
Let $J$ be a bounded interval.
Let $\sequence {f_n}$ be a sequence of real functions $f_n: J \to \R$.
Let each of $\sequence {f_n}$ be differentiable on $J$.
Let $\sequence {\map {f_n} {x_0} }$ be convergent for some $x_0 \in J$.
Let the sequence of derivatives $\sequence { {f_n}'}$ converge uniformly on $J$ to a function $g : J \to \R$.
Then $\sequence {f_n}$ converge uniformly on $J$ to a differentiable function $f: J \to \R$ and $D_x f = g$.
== Proof ==
== Sources ==
* : $\S 8.2.3$
Category:Differential Calculus
Category:Convergence
Category:Continuity
Derivative of Uniformly Convergent Series of Continuously Differentiable Functions

== Theorem ==
Let $\sequence {f_n}$ be a sequence of real functions.
Let each of $\sequence {f_n}$ be continuously differentiable on the interval $\closedint a b$.
Let the series:
:$\ds \map f x := \sum_{n \mathop = 1}^\infty \map {f_n} x$
be pointwise convergent for all $x \in \closedint a b$.
Let the series:
:$\ds \sum_{n \mathop = 1}^\infty \frac \d {\d x} \map {f_n} x$
be uniformly convergent for all $x \in \closedint a b$.
Then:
:$\ds \frac \d {\d x} \map f x := \sum_{n \mathop = 1}^\infty \frac \d {\d x} \map {f_n} x$
== Proof ==
== Sources ==
* : $\S 1.3.1$: Properties of uniformly convergent series: Theorem $1.10$
Category:Differential Calculus
Category:Convergence
Category:Continuity
Category:Series
Derivative of Union is Union of Derivatives

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$, $B$ be subsets of $S$.
Then
:$\paren {A \cup B}' = A' \cup B\,'$
where
:$A'$ denotes the derivative of $A$.
== Proof ==
=== Derivative of Union subset of Union of Derivatives ===
It will be shown that:
:$\paren {A \cup B}' \subseteq A' \cup B\,'$
Let $x \in \paren {A \cup B}'$.
By definition of set derivative:
:$x$ is an accumulation point of $A \cup B$.
 
Then by definition of accumulation point of set:
:$(1): \quad x \in \paren {\paren {A \cup B} \setminus \set x}^-$
where $A^-$ denotes the closure of $A$.
By Set Difference is Right Distributive over Union:
:$\paren {A \cup B} \setminus \set x = \paren {A \setminus \set x} \cup \paren {B \setminus \set x}$
Then by Closure of Finite Union equals Union of Closures:
:$\paren {\paren {A \cup B} \setminus \set x}^- = \paren {A \setminus \set x}^- \cup \paren {B \setminus \set x}^-$
Then by $(1)$ and definition of set union:
:$x \in \paren {A \setminus \set x}^-$ or $x \in \paren {B \setminus \set x}^-$
Then by definition of accumulation point of set:
:$x$ is an accumulation point of $A$ or $x$ is an accumulation point of $B$.
Then by definition of set derivative:
:$x \in A'$ or $x \in B\,'$
Hence by definition of set union:
:$x \in A' \cup B\,'$
=== Union of Derivatives subset of Derivative of Union ===
By Set is Subset of Union:
:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
Then by Derivative of Subset is Subset of Derivative:
:$A' \subseteq \paren {A \cup B}'$ and $B\,' \subseteq \paren {A \cup B}'$
Hence by Union of Subsets is Subset:
:$A' \cup B\,' \subseteq \paren {A \cup B}'$
== Sources ==
* 
* 
Category:Set Derivatives
Derivative of Vector Cross Product of Vector-Valued Functions

== Theorem ==
Let $\mathbf a: \R \to \R^3$ and $\mathbf b: \R \to \R^3$ be differentiable vector-valued functions in Cartesian $3$-space.
The derivative of their vector cross product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b} = \dfrac {\d \mathbf a} {\d x} \times \mathbf b + \mathbf a \times \dfrac {\d \mathbf b} {\d x}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Derivative of Dot Product of Vector-Valued Functions
* Derivative of Product of Real Function and Vector-Valued Function
== Sources ==
* : $\S 22$: Formulas involving Derivatives: $22.24$
Category:Product Rule for Derivatives
Category:Vector Calculus
Category:Vector Cross Product
Category:Derivative of Vector Cross Product of Vector-Valued Functions
Derivative of Vector Cross Product of Vector-Valued Functions/Proof 1

== Theorem ==
== Proof ==
Let:
:$\mathbf a: x \mapsto \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$
:$\mathbf b: x \mapsto \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$
Then:
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 7$
Category:Derivative of Vector Cross Product of Vector-Valued Functions
Derivative of Vector Cross Product of Vector-Valued Functions/Proof 2

== Theorem ==
== Proof ==
Let $\mathbf v = \mathbf a \times \mathbf b$.
Then:
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products: $(3.4)$
Category:Derivative of Vector Cross Product of Vector-Valued Functions
Derivative of Vector Triple Product of Vector-Valued Functions

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be differentiable vector-valued functions in Cartesian $3$-space.
The derivative of their vector triple product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \times \paren {\mathbf b \times \mathbf c} } = \dfrac {\d \mathbf a} {\d x} \times \paren {\mathbf b \times \mathbf c} + \mathbf a \times \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c} + \mathbf a \times \paren {\mathbf b \times \dfrac {\d \mathbf c} {\d x} }$
== Proof ==
== Also see ==
* Derivative of Vector Cross Product of Vector-Valued Functions
* Derivative of Scalar Triple Product of Vector-Valued Functions
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products
Category:Differential Calculus
Category:Vector Calculus
Category:Vector Triple Product
Derivative of x^n by Bessel Function of the First Kind of Order n of x

== Theorem ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
:$\map {\dfrac \d {\d x} } {x^n \map {J_n} x} = x^n \map {J_{n - 1} } x$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {II}$. Bessel functions: $3$
Category:Bessel Functions
Derivative of x to the a x

== Theorem ==
Let $x \in \R$ be a real variable whose domain is the set of (strictly) positive real numbers $\R_{>0}$.
Let $c \in \R_{>0}$ be a (strictly) positive real constant.
Then:
:$\dfrac \d {\d x} x^{a x} = a x^{a x} \paren {\ln x + 1}$
== Proof ==
Let $y := x^{a x}$.
As $x > 0$, we can take the natural logarithm of both sides:
Category:Derivative of x to the x
Derivative of x to the x

== Theorem ==
Let $x \in \R$ be a real variable whose domain is the set of (strictly) positive real numbers $\R_{>0}$.
Then:
:$\dfrac \d {\d x} x^x = x^x \paren {\ln x + 1}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Derivative of x to the x
Category:Derivatives involving Exponential Function
Derivative of x to the x/Proof 1

== Theorem ==
== Proof ==
Note that the Power Rule cannot be used because the index is not a constant.
Let $y := x^x$.
As $x$ was stipulated to be positive, we can take the natural logarithm of both sides:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Variable Index: Example $1$.
* 
Category:Derivative of x to the x
Derivatives of Function of a x + b

== Theorem ==
Let $f$ be a real function which is differentiable on $\R$.
Let $a, b \in \R$ be constants.
Then:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$.
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$.
=== Basis for the Induction ===
$P(1)$ is the case:.
:$\map {\dfrac \d {\d x} } {\map f {a x + b} } = a \map {\dfrac \d {\d z} } {\map f z}$
where $z = a x + b$.
This is proved in Derivative of Function of Constant Multiple: Corollary.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\map {\dfrac {\d^k} {\d x^k} } {\map f {a x + b} } = a^k \map {\dfrac {\d^n} {\d z^k} } {\map f z}$
where $z = a x + b$.
Then we need to show:
:$\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\map f {a x + b} } = a^{k + 1} \map {\dfrac {\d^{k + 1} } {\d z^{k + 1} } } {\map f z}$
where $z = a x + b$.
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$.
Category:Differential Calculus
Category:Derivatives
Derivatives of Hyperbolic Functions

== Theorem ==
This page gathers together derivatives of hyperbolic functions.
Let $u$ be a differentiable real function of $x$.
=== Derivative of Hyperbolic Sine Function ===
=== Derivative of Hyperbolic Cosine Function ===
=== Derivative of Hyperbolic Tangent Function ===
=== Derivative of Hyperbolic Cotangent Function ===
=== Derivative of Hyperbolic Secant Function ===
=== Derivative of Hyperbolic Cosecant Function ===
Category:Derivatives
Category:Derivatives of Hyperbolic Functions
Derivatives of Inverse Trigonometric Functions

== Theorem ==
This page gathers together derivatives of inverse trigonometric functions.
=== Derivative of Arcsine Function ===
=== Derivative of Arccosine Function ===
=== Derivative of Arctangent Function ===
=== Derivative of Arccotangent Function ===
=== Derivative of Arcsecant Function ===
=== Derivative of Arccosecant Function ===
Category:Derivatives
Category:Derivatives of Inverse Trigonometric Functions
Derivatives of Moment Generating Function of Gamma Distribution

== Theorem ==
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Let $t < \beta$.
Let $M_X$ denote the moment generating function of $X$.
The $n$th derivative of $M_X$ is given by:
:${M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$
where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:${M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:${M_X}^{\paren k} = \dfrac {\alpha^{\overline k} \beta^\alpha} {\paren {\beta - t}^{\alpha + k} }$
from which it is to be shown that:
:${M_X}^{\paren {k + 1} } = \dfrac {\alpha^{\overline {k + 1} } \beta^\alpha} {\paren {\beta - t}^{\alpha + k + 1} }$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: {M_X}^{\paren n} = \dfrac {\alpha^{\overline n} \beta^\alpha} {\paren {\beta - t}^{\alpha + n} }$
Category:Moment Generating Function of Gamma Distribution