nyuuzyou/wikis huggingface dataset:
Derivative of Exponential Function/Proof 5

== Theorem ==
== Proof ==
This proof assumes the limit definition of $\exp$.
So let:
:$\forall n \in \N: \forall x \in \R: \map {f_n} x = \paren {1 + \dfrac x n}^n$
Let $x_0 \in \R$.
Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.
Let:
:$N = \ceiling {\max \set {\size {x_0 - 1}, \size {x_0 + 1} } }$
where $\ceiling {\, \cdot \,}$ denotes the ceiling function.
From Closed Real Interval is Compact in Metric Space, $I$ is compact.
From Chain Rule for Derivatives:
:$\dfrac \d {\d x} \map {f_n} x = \dfrac n {n + x} \map {f_n} x$
=== Lemma ===
From the lemma:
:$\forall x \in I: \sequence {\dfrac \d {\d x} \map {f_{n + N} } x}$ is increasing
Hence, from Dini's Theorem, $\sequence {\dfrac \d {\d x} f_{n + N} }$ is uniformly convergent on $I$.
Therefore, for $x \in I$:
In particular:
:$\dfrac \d {\d x} \exp x_0 = \exp x_0$
Category:Derivative of Exponential Function
Derivative of Exponential Function/Proof 5/Lemma

== Theorem ==
For $x \in\ R$, let $\ceiling x$ denote the ceiling of $x$.
Then:
:$\forall x \in \R : n \ge \ceiling {\size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ is increasing.
== Proof ==
First:
Then:
So, for $n \ge \ceiling {\size x}$:
:$\sequence {\dfrac n {n + x} }$
and:
:$\sequence {\paren {1 + \dfrac x n}^n}$
are positive.
Now let $n \ge \ceiling {\size x}$.
Suppose first that $x \in \R_{\ge 0}$.
Then:
So $\sequence {\dfrac n {n + x} }$ is increasing.
Further, from Exponential Sequence is Eventually Increasing:
:$\sequence {\paren {1 + \dfrac x n}^n}$ is increasing.
Hence, from Product of Positive Increasing Functions is Increasing:
:$n \ge \ceiling {size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ is increasing.
Suppose instead that $x \in \R_{<0}$.
:
:$\sequence {\dfrac n {n + x} \sequence {1 + \dfrac x n}^n}$ is decreasing.
From above: $\sequence {1 + \dfrac x n} = \sequence {\dfrac {n + x} n}$ is decreasing.
Thus, from Product of Positive Increasing and Decreasing Functions is Decreasing:
:$\sequence {\dfrac {n + x} n \dfrac n {n + x} \paren {1 + \dfrac x n}^n} = \sequence {\paren {1 + \dfrac x n}^n}$ is decreasing.
This contradicts Exponential Sequence is Eventually Increasing.
Hence the result, by Proof by Contradiction.
Category:Derivative of Exponential Function
Derivative of Exponential Integral Function

== Theorem ==
Let $\Ei: \R_{>0} \to \R$ denote the exponential integral function:
:$\map \Ei x = \ds \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {e^{-t} } t \rd t$
Then:
:$\dfrac \d {\d x} \paren {\map \Ei x} = -\dfrac {e^{-x} } x$
== Proof ==
Category:Derivatives
Category:Exponential Integral Function
Derivative of Exponential at Zero

== Theorem ==
Let $\exp x$ be the exponential of $x$ for real $x$.
Then:
:$\ds \lim_{x \mathop \to 0} \frac {\exp x - 1} x = 1$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Derivatives involving Exponential Function
Category:Derivative of Exponential at Zero
Derivative of Exponential at Zero/Proof 1

== Theorem ==
== Proof ==
For all $x \in \R$:
:$\exp 0 - 1 = 0$ from Exponential of Zero
:$\map {D_x} {\exp x - 1} = \exp x$ from Sum Rule for Derivatives
:$D_x x = 1$ from Derivative of Identity Function.
Its prerequisites having been verified, Corollary 1 to L'Hôpital's Rule yields immediately:
:$\ds \lim_{x \mathop \to 0} \frac {\exp x - 1} x = \lim_{x \mathop \to 0} \frac {\exp x} 1 = \exp 0 = 1$
== Sources ==
* : $\S 14.5 \ (3) \ \text{(i)}$
Category:Derivative of Exponential at Zero
Derivative of Exponential at Zero/Proof 2

== Theorem ==
== Proof ==
Note that this proof does not presuppose Derivative of Exponential Function.
We use the definition of the exponential as a limit of a sequence:
The right summand converges to zero as $h \to 0$, and so:
:$\ds \lim_{h \mathop \to 0} \frac {\exp h - 1} h = 1$
Category:Derivative of Exponential at Zero
Derivative of Exponential of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Let $e^u$ be the exponential function of $u$.
Then:
:$\map {\dfrac \d {\d x} } {e^u} = e^u \dfrac {\d u} {\d x}$
== Proof ==
== Sources ==
* : $\S 13$: Derivatives of Exponential and Logarithmic Functions: $13.29$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $12$
Category:Derivatives involving Exponential Function
Derivative of Fresnel Cosine Integral Function

== Theorem ==
:$\dfrac {\d \mathrm C} {\d x} = \sqrt {\dfrac 2 \pi} \cos x^2$
where $\mathrm C$ denotes the Fresnel cosine integral function.
== Proof ==
We have, by the definition of the Fresnel cosine integral function:
:$\ds \map {\mathrm C} x = \sqrt {\dfrac 2 \pi} \int_0^x \cos u^2 \rd u$
By Fundamental Theorem of Calculus (First Part): Corollary, we therefore have:
:$\dfrac {\d \mathrm C} {\d x} = \sqrt {\dfrac 2 \pi} \cos x^2$
Category:Fresnel Cosine Integral Function
Category:Derivatives
Derivative of Fresnel Sine Integral Function

== Theorem ==
:$\dfrac {\d \mathrm S} {\d x} = \sqrt {\dfrac 2 \pi} \sin x^2$
where $\mathrm S$ denotes the Fresnel sine integral function.
== Proof ==
We have, by the definition of the Fresnel sine integral function:
:$\ds \map {\mathrm S} x = \sqrt {\frac 2 \pi} \int_0^x \sin u^2 \rd u$
By Fundamental Theorem of Calculus (First Part): Corollary, we therefore have:
:$\dfrac {\d \mathrm S} {\d x} = \sqrt {\dfrac 2 \pi} \sin x^2$
Category:Fresnel Sine Integral Function
Category:Derivatives
Derivative of Function of Constant Multiple

== Theorem ==
Let $f$ be a real function which is differentiable on $\R$.
Let $c \in \R$ be a constant.
Then:
:$\map {D_x} {\map f {c x} } = c \map {D_{c x} } {\map f {c x} }$
=== Corollary ===
== Proof ==
First it is shown that $\map {D_x} {c x} = c$:
Next:
== Examples ==
Category:Derivative of Function of Constant Multiple
Category:Derivative of Constant Multiple
Category:Derivatives
Derivative of Function of Constant Multiple/Corollary

== Corollary to Derivative of Function of Constant Multiple ==
Let $f$ be a real function which is differentiable on $\R$.
Let $a, b \in \R$ be constants.
Then:
:$\map {\dfrac \d {\d x} } {\map f {a x + b} } = a \, \map {\dfrac \d {\map \d {a x + b} } } {\map f {a x + b} }$
== Proof ==
First it is shown that $\map {\dfrac \d {\d x} } {a x + b} = a$:
Next:
Category:Derivative of Function of Constant Multiple
Category:Derivative of Constant Multiple
Derivative of Function of Constant Multiple/Examples/Cosine of a x + b

== Example of Derivative of Function of Constant Multiple ==
:$\map {\dfrac \d {\d x} } {\map \cos {a x + b} } = -a \map \sin {a x + b}$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $9$.
Category:Examples of Derivative of Function of Constant Multiple
Derivative of Function of Constant Multiple/Examples/Secant of a x + b

== Example of Derivative of Function of Constant Multiple ==
:$\map {\dfrac \d {\d x} } {\map \sec {a x + b} } = a \map \sec {a x + b} \map \tan {a x + b}$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $20$.
Category:Examples of Derivative of Function of Constant Multiple
Derivative of Function of Constant Multiple/Examples/Sine 2 x

== Example of Derivative of Function of Constant Multiple ==
:$\map {\dfrac \d {\d x} } {\sin 2 x} = 2 \cos 2 x$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $7$.
Category:Examples of Derivative of Function of Constant Multiple
Derivative of Function plus Constant

== Theorem ==
Let $f$ be a real function which is differentiable on $\R$.
Let $c \in \R$ be a constant.
Then:
:$\map {\dfrac \d {\d x} } {\map f x + c} = \map {\dfrac \d {\d x} } {\map f x}$
== Proof ==
Category:Differential Calculus
Derivative of Function to Power of Function

== Theorem ==
Let $\map u x, \map v x$ be real functions which are differentiable on $\R$.
Then:
:$\map {\dfrac \d {\d x} } {u^v} = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v$
== Proof ==
== Also presented as ==
== Also see ==
* Power Rule for Derivatives: when $u = x$ and $v = n$ where $n$ is constant:
:$\map {\dfrac \d {\d x} } {x^n} = n x^{n-1}$
* Derivative of Exponential Function: when $v = x$ and $u = a$ where $a$ is constant:
:$\map {\dfrac \d {\d x} } {a^x} = a^x \ln a$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Derivatives: $3.3.6$
* : $\S 13$: Derivatives of Exponential and Logarithmic Functions: $13.30$
Category:Derivative of Function to Power of Function
Category:Differential Calculus
Category:Exponential Function
Derivative of Function to Power of Function/Also presented as

== Derivative of Function to Power of Function: Also presented as==
Derivative of Function to Power of Function can also be seen presented in the form:
:$\map {\dfrac \d {\d x} } {u^v} = u^v \paren {\dfrac v u \dfrac {\d u} {\d x} + \ln u \dfrac {\d v} {\d x} }$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Derivatives: $3.3.6$
Category:Derivative of Function to Power of Function
Derivative of Gamma Function

== Theorem ==
:$\ds \map {\Gamma'} x = \int_0^\infty t^{x - 1} \ln t \, e^{-t} \rd t$
where $\map {\Gamma'} x$ denotes the derivative of the Gamma function evaluated at $x$. 
== Proof == 
Category:Gamma Function
Category:Derivatives
Derivative of Gamma Function at 1

== Theorem ==
Let $\Gamma$ denote the Gamma function.
Then:
:$\map {\Gamma'} 1 = \ds \int_0^\infty e^{-x} \ln x \rd x = -\gamma$
where:
:$\map {\Gamma'} 1$ denotes the derivative of the Gamma function evaluated at $1$
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 16$: Derivatives of the Gamma Function: $16.13$
* : $0,57721 56649 \ldots$
* : $\S 25$: The Gamma Function: Derivatives of the Gamma Function: $25.13.$
Category:Derivative of Gamma Function at 1
Category:Gamma Function
Category:Euler-Mascheroni Constant
Derivative of Gamma Function at 1/Proof 1

== Theorem ==
== Proof ==
From Reciprocal times Derivative of Gamma Function:
:$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$
Setting $z = 1$:
Using Gamma Function Extends Factorial:
:$\map \Gamma 1 = \paren {1 - 1}! = 1$
Hence:
:$\map {\Gamma'} 1 = -\gamma \map \Gamma 1 = -\gamma$
Category:Derivative of Gamma Function at 1
Derivative of Gaussian Hypergeometric Function

== Theorem ==
:$\map {\dfrac \d {\d x} } {\map F {a, b; c; x} } = \dfrac {a b} c \map F {a + 1, b + 1; c + 1; x} $
where:
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
== Proof ==
== Also see ==
* Gauss's Hypergeometric Theorem
== Sources ==
* : $\S 31$: Hypergeometric Functions: Miscellaneous Properties: $31.15$
Category:Gaussian Hypergeometric Function
Category:Derivatives
Derivative of General Exponential Function

== Theorem ==
Let $a \in \R_{>0}$.
Let $a^x$ be $a$ to the power of $x$.
Then:
:$\map {\dfrac \d {\d x} } {a^x} = a^x \ln a$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
Category:Derivative of General Exponential Function
Category:Derivatives involving Exponential Function
Derivative of General Logarithm Function

== Theorem ==
Let $a \in \R_{>0}$ such that $a \ne 1$
Let $\log_a x$ be the logarithm function to base $a$.
Then:
:$\map {\dfrac \d {\d x} } {\log_a x} = \dfrac {\log_a e} x$
== Proof ==
== Also presented as ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation
Category:Derivative of General Logarithm Function
Category:Derivatives involving Logarithm Functions
Category:Logarithms
Derivative of General Logarithm Function/Also presented as

== Derivative of General Logarithm Function: Also presented as ==
This result can also be seen presented as:
:$\map {\dfrac \d {\d x} } {\log_a x} = \dfrac 1 {x \ln a}$
where $\ln a := \log_e a$ is the natural logarithm of $a$.
This is seen to be equivalent to the given form by Logarithm of Reciprocal:
:$\dfrac 1 {\ln a} = \log_a e$
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Log to base $a$
Category:Derivative of General Logarithm Function
Derivative of General Logarithm of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Let $a \in \R_{>0}$ such that $a \ne 1$
Let $\log_a u$ be the logarithm to base $a$ of $u$.
Then:
:$\map {\dfrac \d {\d x} } {\log_a u} = \dfrac {\log_a e} u \dfrac {\d u} {\d x}$
== Proof ==
== Sources ==
* : $\S 13$: Derivatives of Exponential and Logarithmic Functions: $13.26$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $9$
Category:Derivative of General Logarithm Function
Derivative of Generating Function/General Result

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $m$ be a positive integer.
Then:
:$\ds \dfrac {\d^m} {\d z^m} \map G z = \sum_{k \mathop \ge 0} \dfrac {\paren {k + m}!} {k!} a_{k + m} z^k$
=== Corollary ===
== Proof ==
Proof by induction:
=== Basis for the Induction ===
When $n = 0$, we have:
:$\dfrac {\d^0} {\d z^0} \map G z = \map G z$
Also:
:$\ds \sum_{k \mathop \ge 0} \dfrac {\paren {k + 0}!} {k!} a_{k + 0} z^k = \sum_{k \mathop \ge 0} a_k z^k$
This is the basis for the induction.
=== Induction Hypothesis ===
:$\ds \forall n \in \N: n \ge 0: \dfrac {\d^n} {\d z^n} \map G z = \sum_{k \mathop \ge 0} \dfrac {\paren {k + n}!} {k!} a_{k + n} z^k$
This is the induction hypothesis.
It is to be demonstrated that:
:$\ds \dfrac {\d^{n + 1} } {\d z^{n + 1} } \map G z = \sum_{k \mathop \ge 0} \dfrac {\paren {k + n + 1}!} {k!} a_{k + n + 1} z^k$
=== Induction Step ===
This is the induction step:
Consider $m = n + 1$.
Hence the result by induction.
Category:Generating Functions
Derivative of Generating Function/General Result/Corollary

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $m$ be a positive integer.
Let the coefficient of $z^n$ extracted from $\map G z$ be denoted:
:$\sqbrk {z^n} \map G z := a_n$
Then:
:$\sqbrk {z^m} \map G z = \dfrac 1 {m!} \map {G^{\paren m} } 0$
where $G^{\paren m}$ denotes the $m$th derivative of $G$.
== Proof ==
Category:Generating Functions
Derivative of Generating Function for Sequence of Harmonic Numbers

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = H_n$
where $H_n$ denotes the $n$th harmonic number.
Let $\map G z$ be the generating function for $\sequence {a_n}$:
:$\map G z = \dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} }$
from Generating Function for Sequence of Harmonic Numbers.
Then the derivative of $\map G z$ $z$ is given by:
:$\map {G'} z = \dfrac 1 {\paren {1 - z}^2} \map \ln {\dfrac 1 {1 - z} } + \dfrac 1 {\paren {1 - z}^2}$
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $3$
Category:Generating Functions
Category:Harmonic Numbers
Derivative of Geometric Sequence

== Theorem ==
Let $x \in \R: \size x < 1$.
Then:
:$\ds \sum_{n \mathop \ge 1} n x^{n - 1} = \frac 1 {\paren {1 - x}^2}$
=== Corollary ===
== Proof ==
We have from Power Rule for Derivatives that:
:$\ds \frac \d {\d x} \sum_{n \mathop \ge 1} x^n = \sum_{n \mathop \ge 1} n x^{n - 1}$
But from Sum of Infinite Geometric Sequence: Corollary:
:$\ds \sum_{n \mathop \ge 1} x^n = \frac x {1 - x}$
The result follows by Power Rule for Derivatives and the Chain Rule for Derivatives applied to $\dfrac x {1 - x}$.
== Sources ==
* : $\S 2.4$: Expectation: Example $24$: Footnote
Category:Analysis
Category:Differential Calculus
Category:Geometric Sequences
Derivative of Geometric Sequence/Corollary

== Theorem ==
Let $x \in \R: \size x < 1$.
Then:
:$\ds \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1} = \frac 2 {\paren {1 - x}^3}$
== Proof ==
We have from Power Rule for Derivatives that:
:$\ds \frac {\d} {\d x} \sum_{n \mathop \ge 1} \paren {n + 1} x^n = \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1}$
But from Sum of Infinite Geometric Sequence:
The result follows by Power Rule for Derivatives and the Chain Rule for Derivatives applied to $\dfrac 1 {\paren {1 - x}^2}$.
Category:Analysis
Category:Differential Calculus
Category:Geometric Sequences
Derivative of Hyperbolic Cosecant

== Theorem ==
:$\map {\dfrac \d {\d x} } {\csch x} = -\csch x \coth x$
where:
:$\coth x$ denotes the hyperbolic cotangent and $\csch z$ denotes the hyperbolic cosecant.
:$x \in \R_{\ne 0}$
== Proof ==
It is noted that at $x = 0$, $\csch x$ is undefined.
Hence the restriction of the domain.
== Also see ==
* Derivative of Hyperbolic Sine
* Derivative of Hyperbolic Cosine
* Derivative of Hyperbolic Tangent
* Derivative of Hyperbolic Cotangent
* Derivative of Hyperbolic Secant
== Sources ==
* : Appendix $2$: Table of derivatives and integrals of common functions: Hyperbolic functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Cosecant Function
Derivative of Hyperbolic Cosecant Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\csch u} = -\csch u \coth u \dfrac {\d u} {\d x}$
where $\coth x$ denotes the hyperbolic cotangent and $\csch x$ denotes the hyperbolic cosecant.
== Proof ==
== Also see ==
* 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
== Sources ==
* : $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.36$
* : Chapter $15$: Differentiation of Hyperbolic Functions: Differentiation Formulas: $36$.
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $24$
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Cosecant Function
Derivative of Hyperbolic Cosecant of a x

== Theorem ==
:$\map {\dfrac \d {\d x} } {\csch a x} = -a \csch a x \coth a x$
== Proof ==
== Also see ==
* Derivative of $\csch a x$
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Cosecant Function
Derivative of Hyperbolic Cosine

== Theorem ==
:$\map {\dfrac \d {\d x} } {\cosh x} = \sinh x$
where $\cosh$ is the hyperbolic cosine and $\sinh$ is the hyperbolic sine.
== Proof ==
== Also see ==
* Derivative of Hyperbolic Sine
* Derivative of Hyperbolic Tangent
* Derivative of Hyperbolic Cotangent
* Derivative of Hyperbolic Secant
* Derivative of Hyperbolic Cosecant
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $9$.
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $6.$ Hyperbolic trigonometric functions
* 
* : Appendix $2$: Table of derivatives and integrals of common functions: Hyperbolic functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
* 
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Cosine Function
Derivative of Hyperbolic Cosine Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\cosh u} = \sinh u \dfrac {\d u} {\d x}$
where $\cosh$ is the hyperbolic cosine and $\sinh$ is the hyperbolic sine.
== Proof ==
== Also see ==
* Derivative of Hyperbolic Sine Function
* Derivative of Hyperbolic Tangent Function
* Derivative of Hyperbolic Cotangent Function
* Derivative of Hyperbolic Secant Function
* Derivative of Hyperbolic Cosecant Function
== Sources ==
* : $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.32$
* : Chapter $15$: Differentiation of Hyperbolic Functions: Differentiation Formulas: $32$.
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $20$
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Cosine Function
Derivative of Hyperbolic Cotangent

== Theorem ==
:$\map {\dfrac \d {\d x} } {\coth x} = -\csch^2 x = \dfrac {-1} {\sinh^2 x}$
where $\coth x$ denotes the hyperbolic cotangent, $\csch x$ denotes the hyperbolic cosecant and $\sinh x$ denotes the hyperbolic sine.
=== Corollary ===
== Proof ==
== Also see ==
* Derivative of Hyperbolic Sine
* Derivative of Hyperbolic Cosine
* Derivative of Hyperbolic Tangent
* Derivative of Hyperbolic Secant
* Derivative of Hyperbolic Cosecant
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $11$.
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $6.$ Hyperbolic trigonometric functions
* : Appendix $2$: Table of derivatives and integrals of common functions: Hyperbolic functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
* 
Category:Derivative of Hyperbolic Cotangent
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Cotangent Function
Derivative of Hyperbolic Cotangent/Corollary

== Theorem ==
:$\map {\dfrac \d {\d x} } {\coth x} = 1 - \coth^2 x$
where $\coth x$ denotes the hyperbolic cotangent.
== Proof ==
== Sources ==
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $6.$ Hyperbolic trigonometric functions
Category:Derivative of Hyperbolic Cotangent
Derivative of Hyperbolic Cotangent Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\coth u} = -\csch^2 u \dfrac {\d u} {\d x}$
where $\coth x$ denotes the hyperbolic cotangent and $\csch x$ denotes the hyperbolic cosecant.
== Proof ==
== Also see ==
* Derivative of Hyperbolic Sine Function
* Derivative of Hyperbolic Cosine Function
* Derivative of Hyperbolic Tangent Function
* Derivative of Hyperbolic Secant Function
* Derivative of Hyperbolic Cosecant Function
== Sources ==
* : $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.34$
* : Chapter $15$: Differentiation of Hyperbolic Functions: Differentiation Formulas: $34$.
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $22$
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Cotangent Function
Derivative of Hyperbolic Secant

== Theorem ==
:$\map {\dfrac \d {\d x} } {\sech x} = -\sech x \tanh x$
where $\tanh$ is the hyperbolic tangent and $\sech$ is the hyperbolic secant.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Derivative of Hyperbolic Sine
* Derivative of Hyperbolic Cosine
* Derivative of Hyperbolic Tangent
* Derivative of Hyperbolic Cotangent
* Derivative of Hyperbolic Cosecant
== Sources ==
* : Appendix $2$: Table of derivatives and integrals of common functions: Hyperbolic functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Secant Function
Category:Derivative of Hyperbolic Secant Function
Derivative of Hyperbolic Secant Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\sech u} = -\sech u \tanh u \dfrac {\d u} {\d x}$ 
where $\tanh$ is the hyperbolic tangent and $\sech$ is the hyperbolic secant.
== Proof ==
== Also see ==
* Derivative of Hyperbolic Sine Function
* Derivative of Hyperbolic Cosine Function
* Derivative of Hyperbolic Tangent Function
* Derivative of Hyperbolic Cotangent Function
* Derivative of Hyperbolic Cosecant Function
== Sources ==
* : $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.35$
* : Chapter $15$: Differentiation of Hyperbolic Functions: Differentiation Formulas: $35$.
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $23$
Category:Derivatives of Hyperbolic Functions
Category:Hyperbolic Secant Function
Category:Derivative of Hyperbolic Secant Function