nyuuzyou/wikis huggingface dataset:
Derivative of Natural Logarithm Function/Examples/(x-2)^1/3 times (x-3)^1/2 times (2x-1)^3/2/Mistake

== Source Work ==
:
:$\text {II}$. Calculus: Differentiation: Variable Index: Example
== Mistake ==
:Let $y = \paren {x - 2}^{1/3} \paren {x - 3}^{1/2} \paren {2 x - 1}^{3/2}$,
:then $\dfrac {\d y} {\d x} = \paren {\dfrac 1 {3 \paren {x - 2} } + \dfrac 1 {2 \paren {x - 3} } + \dfrac 3 {2 x - 1} } \paren {x - 1}^{1/3} \paren {x - 3}^{1/2} \paren {2 x - 1}^{3/2}$
== Correction ==
Should be:
:$\dfrac {\d y} {\d x} = \paren {\dfrac 1 {3 \paren {x - 2} } + \dfrac 1 {2 \paren {x - 3} } + \dfrac 3 {2 x - 1} } \paren {x - 2}^{1/3} \paren {x - 3}^{1/2} \paren {2 x - 1}^{3/2}$
See Derivative of $\paren {x - 2}^{1/3} \paren {x - 3}^{1/2} \paren {2 x - 1}^{3/2}$.
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Variable Index: Example
Category:Mistakes/A Note Book in Pure Mathematics
Derivative of Natural Logarithm Function/Proof 2

== Theorem ==
== Proof ==
This proof assumes the definition of the natural logarithm as the inverse of the exponential function, where the exponential function is defined as the limit of a sequence:
:$e^x := \ds \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$
It also assumes the Laws of Logarithms.
Define $u$ as:
Hence:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Standard Differential Coefficients
* 
Category:Derivative of Natural Logarithm Function
Derivative of Natural Logarithm Function/Proof 3

== Theorem ==
== Proof ==
This proof assumes the definition of the natural logarithm as the inverse of the exponential function as defined by differential equation:
:$y = \dfrac {\d y} {\d x}$
:$y = e^x \iff \ln y = x$
The result follows from the definition of the antiderivative and the defined initial condition:
:$\tuple {x_0, y_0} = \tuple {0, 1}$
Category:Derivative of Natural Logarithm Function
Derivative of Natural Logarithm Function/Proof 4

== Theorem ==
== Proof ==
This proof assumes the definition of the natural logarithm as the limit of a sequence of real functions.
Let $\sequence {f_n}$ be the sequence of mappings $f_n: \R_{>0} \to \R$ defined as:
:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Fix $x_0 \in \R_{>0}$.
Pick $k \in \N : x_0 \in J := \closedint {\dfrac 1 k} k$.
From definition of bounded interval, $J$ is bounded.
From Derivative of Nth Root and Combination Theorem for Sequences:
:$\forall n \in \N : \forall x \in J : D_x \map {f_n} x = \dfrac {\sqrt [n] x} x$
In particular:
:$\forall n: f_n$ is differentiable on $J$
From Defining Sequence of Natural Logarithm is Convergent, $\sequence {\map {f_n} {x_0} }$ is convergent.
=== Lemma ===
From the lemma, $\sequence { {f_n}'}$ converges uniformly to $\dfrac 1 x$ on $J$.
From Derivative of Uniformly Convergent Sequence of Differentiable Functions, $\map {f'} x = \dfrac 1 x$ on $J$
In particular:
:$\map {f'} {x_0} = \dfrac 1 {x_0}$
Hence the result.
Category:Derivative of Natural Logarithm Function
Derivative of Natural Logarithm Function/Proof 4/Lemma

== Theorem ==
Let $\sequence {f_n}_n$ be the sequence of real functions $f_n: \R_{>0} \to \R$ defined as:
:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Let $k \in \N$.
Let $J = \closedint {\dfrac 1 k} k$.
Then the sequence of derivatives $\sequence { {f_n}'}_n$ converges uniformly to some real function $g: J \to \R$.
== Proof ==
From Derivative of $n$th Root and Combination Theorem for Sequences:
:$\forall n \in \N: \forall x \in J : D_x \map {f_n} x = \dfrac {\sqrt [n] x} x$
From Closed Bounded Subset of Real Numbers is Compact, $J$ is compact.
Thus:
In particular, $\sequence { {f_n}'}_n$ is pointwise convergent to a continuous function on $J$.
For $x \in \closedint {\frac 1 k} 1$:
So $\sequence {\map { {f_n}'} x}_n$ is increasing when $x \in \closedint {\dfrac 1 k} 1$.
For $x \in \closedint 1 k$:
So $\sequence {\map { {f_n}'} x}_n$ is decreasing when $x \in \closedint {\dfrac 1 k} 1$.
Thus $\sequence {\map { {f_n}'} x}_n$ is monotone for all $x \in \closedint {\frac 1 k} 1 \cup \closedint 1 k = J$.
From Dini's Theorem, $\sequence { {f_n}'}_n$ converges uniformly to $\dfrac 1 x$ on $J$.
Hence the result.
Category:Derivative of Natural Logarithm Function
Derivative of Natural Logarithm of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Let $\ln u$ be the natural logarithm of $u$.
Then:
:$\map {\dfrac \d {\d x} } {\ln u} = \dfrac 1 u \dfrac {\d u} {\d x}$
== Proof ==
== Sources ==
* : $\S 13$: Derivatives of Exponential and Logarithmic Functions: $13.27$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $10$
Category:Derivative of Natural Logarithm Function
Derivative of Nth Root

== Theorem ==
Let $n \in \N_{>0}$.
Let $f: \R \to \R$ be the real function defined as $\map f x = \sqrt [n] x$.
Then:
:$\map {f'} x = \dfrac 1 {n \paren {\sqrt [n] x}^{n - 1} }$
everywhere that $\map f x = \sqrt [n] x$ is defined.
== Proof ==
Category:Derivatives
Derivative of P-Norm wrt P

== Theorem ==
Let $p \ge 1$ be a real number.
Let $\ell^p$ denote the $p$-sequence space.
Let $\mathbf x = \sequence {x_n} \in \ell^p$.
Let $\norm {\mathbf x}_p$ be a $p$-norm.
Suppose, $\norm {\mathbf x}_p \ne 0$.
Then:
:$\ds \dfrac \d {\d p} \norm {\mathbf x}_p = \frac {\norm {\mathbf x}_p} p \paren { \frac {\sum_{n \mathop = 0}^\infty \size {x_n}^p \map \ln {\size {x_n} } } {\norm {\bf x}_p^p} - \map \ln {\norm {\bf x}_p} }$
== Proof ==
We begin with the natural logarithm of $\norm {\mathbf x}_p$:
Multiplication by $\norm {\mathbf x}_p$ completes the proof.
Category:P-Norms
Derivative of Periodic Real Function

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $f$ be differentiable on all of $\R$.
Let $f$ be periodic with period $L$.
Then its derivative is also periodic with period $L$.
== Proof ==
Let $f$ be differentiable on all of $\R$.
Let $f$ be periodic with period $L$.
Then taking the derivative of both sides using the Chain Rule for Derivatives yields:
:$\map f x = \map f {x + L} \implies \map {f'} x = \map {f'} {x + L}$
Let $L'$ be the period of $f'$.
$f$ is differentiable and therefore continuous, by Differentiable Function is Continuous.
From Image of Closed Real Interval is Bounded, it follows that $f$ is bounded on $\closedint 0 {\size L}$.
But from the General Periodicity Property, it follows that $f$ is bounded on all of $\R$.
Then from Primitive of Periodic Real Function, it follows that $L'$ is the period of $f$.
But we had previously established that $L$ was the period of $f$.
Hence it follows that $L' = L$.
Hence the result.
== Also see ==
* Primitive of Periodic Real Function
Category:Differential Calculus
Category:Periodic Functions
Derivative of Power of Function

== Theorem ==
Let $\map u x$ be a differentiable real function of $x$.
Let $n$ be a real number such that $n \ne -1$.
Then:
:$\map {\dfrac \d {\d x} } {\map u x^n} = n \map u x^{n - 1} \map {\dfrac \d {\d x} } {\map u x}$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
This can be (and usually is) presented more simply as:
:$\map {\dfrac \d {\d x} } {u^n} = n u^{n - 1} \dfrac {\d u} {\d x}$
== Sources ==
* : $\S 13$: General Rules of Differentiation: $13.10$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $2$
Category:Derivative of Power of Function
Category:Differential Calculus
Derivative of Product of Operator-Valued Functions

== Theorem ==
Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$, and $\struct {Z, \norm \cdot_Z}$ normed vector spaces.
Let $\map B {X, Y}$, $\map B {Y, Z}$ and $\map B {X, Z} $ denote the space of bounded linear transformations between $X$ and $Y$, between $Y$ and $Z$, and between $X$ and $Z$, respectively.
Let $A : I \to \map B {X, Y}$ and $B : I \to \map B {Y, Z}$ be differentiable mappings defined on an interval $I$ whose images are bounded linear transformations.
The product $A B: I \to \map B {X, Z}: x \mapsto \map A x \map B x$ is differentiable and its derivative at any $x \in \R$ is given by $\map {A'} x \map B x + \map A x \map {B'} x$.
== Proof ==
Given arbitrary $x, h \in I$ such that $h \ne 0$, $x+h\in I$ we compute:
:$\norm {\dfrac {\map {\paren {A B} } {x + h} - \map {\paren {A B} } x} h - \map {A'} x \map B x - \map A x \map {B'} x}_{\map B {X, Z} }$
to be:
$\norm {\dfrac {\map A {x + h} - \map A x} h - \map {A'} x}_{\map B {X, Y} }$ vanishes as $h \to 0$ because $\map {A'} x$ is the derivative of $A$ at $x$ which exists by assumption.
$\norm {\dfrac {\map B {x + h} - \map B x} h - \map {B'} x}_{\map B {Y, Z} }$ vanishes as $h \to 0$ because $\map {B'} x$ is the derivative of $B$ at $x$ which exists by assumption.
$\norm {\map B {x + h} - \map B x }_{\map B {Y, Z} }$ vanishes as $h \to 0$ by Differentiable Operator-Valued Function is Continuous.
By the Reverse Triangle Inequality:
:$\size {\norm {\map B {x + h} }_{\map B {Y, Z} } - \norm {\map B x}_{\map B {Y, Z} } } \le \norm {\map B {x + h} - \map B x}_{\map B {Y, Z} } \to 0$
as $h \to 0$.
Together this shows:
:$\ds \lim_{h \mathop \to 0} \norm {\frac {\map {\paren {A B} } {x + h} - \map {\paren {A B} } x} h - \map {A'} x \map B x - \map A x \map {B'} x}_{\map B {X, Z} } = 0$
Therefore the derivative of $A B$ at $x$ is:
:$\map {A'} x \map B x + \map A x \map {B'} x$
== Sources ==
* : Chapter $\text {IV}$: Banach Algebras: $4.6$. Functions on scalars to the algebra. Equation $(4.6.1)$
Category:Product Rule for Derivatives
Category:Bounded Linear Transformations
Derivative of Product of Real Function and Vector-Valued Function

== Theorem ==
Let:
:$\mathbf z:\R \to \R^n$
be a differentiable vector-valued function, where:
:$\mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix}$
such that:
:$z_1, z_2, \cdots, z_n$
are differentiable real functions.
Let:
:$y: \R \to \R$
be a differentiable real function.
Then:
:$D_x \left({y \, \mathbf z}\right) = \dfrac {\d y} {\d x} \mathbf z + y \dfrac {\d \mathbf z} {\d x}$
== Proof ==
== Also see ==
* Derivative of Vector Cross Product of Vector-Valued Functions
* Derivative of Dot Product of Vector-Valued Functions
Category:Product Rule for Derivatives
Category:Vector Calculus
Derivative of Real Area Hyperbolic Cosecant of x over a

== Theorem ==
:$\dfrac {\map \d {\arcsch \dfrac x a} } {\d x} = \dfrac {-a} {\size x \sqrt {a^2 + x^2} }$
where $x \ne 0$.
== Proof ==
Let $0 < x < a$.
Then $0 < \dfrac x a < 1$ and so:
$\arcsch \dfrac x a$ is not defined when $x = 0$.
== Also see ==
* Derivative of $\arsinh \dfrac x a$
* Derivative of $\arcosh \dfrac x a$
* Derivative of $\artanh \dfrac x a$
* Derivative of $\arcoth \dfrac x a$
* Derivative of $\arsech \dfrac x a$
Category:Derivatives of Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Cosecant
Derivative of Real Area Hyperbolic Cosine

== Theorem ==
Let $x \in \R_{>1}$ be a real number.
Let $\arcosh x$ be the real area hyperbolic cosine of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\arcosh x} = \dfrac 1 {\sqrt {x^2 - 1} }$
== Proof ==
Note that when $y = 0$, $\cosh y$ is defined and equals $1$.
But from Derivative of Hyperbolic Cosine:
:$\valueat {\dfrac \d {\d y} \cosh y} {y \mathop = 0} = \sinh 0 = 0$
Thus $\dfrac {\d y} {\d x} = \dfrac 1 {\sinh y}$ is not defined at $y = 0$.
Hence the limitation of the domain of $\map {\dfrac \d {\d x} } {\arcosh x}$ to exclude $x = 1$.
Now it is necessary to determine the sign of $\dfrac {\d y} {\d x}$.
From:
:Real Area Hyperbolic Cosine is Strictly Increasing
:Derivative of Strictly Increasing Real Function is Strictly Positive
it follows that $\map {\dfrac \d {\d x} } {\arcosh x} > 0$ on $\R_{>1}$.
Thus:
:$\dfrac {\d y} {\d x} = \dfrac 1 {\sqrt {\cosh^2 y - 1} }$
where $\sqrt {\cosh^2 y - 1}$ denotes the positive square root of $\cosh^2 y - 1$.
Hence by definition of $x$ and $y$ above:
:$\map {\dfrac \d {\d x} } {\arcosh x} = \dfrac 1 {\sqrt {x^2 - 1} }$
== Sources ==
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $7.$ Inverse hyperbolic trigonometric functions
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse hyperbolic functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
Category:Derivatives of Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Cosine
Category:Derivative of Real Area Hyperbolic Cosine
Derivative of Real Area Hyperbolic Cosine of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\arcosh u} = \dfrac 1 {\sqrt {u^2 - 1} } \dfrac {\d u} {\d x}$ 
where $u > 1$
where $\cosh^{-1}$ is the real area hyperbolic cosine.
== Proof ==
== Also see ==
* Derivative of Inverse Hyperbolic Sine Function
* Derivative of Inverse Hyperbolic Tangent Function
* Derivative of Inverse Hyperbolic Cotangent Function
* Derivative of Inverse Hyperbolic Secant Function
* Derivative of Inverse Hyperbolic Cosecant Function
== Sources ==
* : $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.38$
* : Chapter $15$: Differentiation of Hyperbolic Functions: Definitions of Inverse Hyperbolic Functions: $38$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $26$
Category:Derivative of Real Area Hyperbolic Cosine
Derivative of Real Area Hyperbolic Cosine of x over a

== Theorem ==
:$\dfrac {\map \d {\map \arcosh {\frac x a} } } {\d x} = \dfrac 1 {\sqrt {x^2 - a^2} }$
where $x > a$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $x > a$.
Then $\dfrac x a > 1$ and so:
$\cosh^{-1} \dfrac x a$ is not defined when $x \le a$.
When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and so $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.
Hence the restriction on the domain.
== Also see ==
* Derivative of $\arsinh \dfrac x a$
* Derivative of $\artanh \dfrac x a$
* Derivative of $\arcoth \dfrac x a$
* Derivative of $\arsech \dfrac x a$
* Derivative of $\arcsch \dfrac x a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $15$.
Category:Derivative of Real Area Hyperbolic Cosine
Derivative of Real Area Hyperbolic Cosine of x over a/Corollary 1

== Theorem ==
:$\map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 - a^2} } } = \dfrac 1 {\sqrt {x^2 - a^2} }$
for $x > a$.
== Proof ==
When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.
When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.
When $x < -a$ we have that $x + \sqrt {x^2 - a^2} < 0$ and so $\map \ln {x + \sqrt {x^2 - a^2} }$ is not defined.
Hence the restriction on the domain.
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $15$.
Category:Derivative of Real Area Hyperbolic Cosine
Derivative of Real Area Hyperbolic Cosine of x over a/Corollary 2

== Theorem ==
:$\map {\dfrac \d {\d x} } {\map \ln {x - \sqrt {x^2 - a^2} } } = -\dfrac 1 {\sqrt {x^2 - a^2} }$
for $x > a$.
== Proof ==
When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.
When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.
When $x < -a$ we have that $x - \sqrt {x^2 - a^2} < 0$ and so $\map \ln {x + \sqrt {x^2 - a^2} }$ is not defined.
Hence the restriction on the domain.
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $15$.
Category:Derivative of Real Area Hyperbolic Cosine
Derivative of Real Area Hyperbolic Cotangent of x over a

== Theorem ==
:$\map {\dfrac \d {\d x} } {\map \arcoth {\dfrac x a} } = \dfrac {-a} {x^2 - a^2}$
where $x^2 > a^2$.
=== Corollary ===
== Proof ==
Let $x^2 > a^2$.
Then either $\dfrac x a < -1$ or $\dfrac x a > 1$ and so:
$\arcoth \dfrac x a$ is not defined when $x^2 \le a^2$.
== Also presented as ==
This result can also be (and usually is) reported as:
:$\map {\dfrac {\d} {\d x} } {\map \arcoth {\dfrac x a} } = \dfrac a {a^2 - x^2}$
but this obscures the fact that $x^2 > a^2$ in order for it to be defined.
Some sources present it as:
:$\map {\dfrac {\d} {\d x} } {\dfrac 1 a \map \arcoth {\dfrac x a} } = \dfrac 1 {a^2 - x^2}$
== Also see ==
* Derivative of $\arsinh \dfrac x a$
* Derivative of $\arcosh \dfrac x a$
* Derivative of $\artanh \dfrac x a$
* Derivative of $\arsech \dfrac x a$
* Derivative of $\arcsch \dfrac x a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
Category:Derivative of Inverse Hyperbolic Cotangent
Derivative of Real Area Hyperbolic Cotangent of x over a/Corollary

== Theorem ==
:$\map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } } = \dfrac 1 {a^2 - x^2}$
where $\size x > a$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
Category:Derivative of Inverse Hyperbolic Cotangent
Derivative of Real Area Hyperbolic Secant of x over a

== Theorem ==
:$\dfrac {\map \d {\map \arsech {\frac x a} } } {\d x} = \dfrac {-a} {x \sqrt{a^2 - x^2} }$
where $0 < x < a$.
== Proof ==
Let $0 < x < a$.
Then $0 < \dfrac x a < 1$ and so:
$\arsech \dfrac x a$ is not defined when $x \le 0$ or $x \ge a$.
== Also see ==
* Derivative of $\arsinh \dfrac x a$
* Derivative of $\arcosh \dfrac x a$
* Derivative of $\artanh \dfrac x a$
* Derivative of $\arcoth \dfrac x a$
* Derivative of $\arcsch \dfrac x a$
Category:Derivatives of Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Secant
Derivative of Real Area Hyperbolic Sine of x over a

== Theorem ==
:$\dfrac {\map \d {\map \arsinh {\frac x a} } } {\d x} = \dfrac 1 {\sqrt {x^2 + a^2}}$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
== Also see ==
* Derivative of $\arcosh \dfrac x a$
* Derivative of $\artanh \dfrac x a$
* Derivative of $\arcoth \dfrac x a$
* Derivative of $\arsech \dfrac x a$
* Derivative of $\arcsch \dfrac x a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $14$.
Category:Derivative of Inverse Hyperbolic Sine
Derivative of Real Area Hyperbolic Sine of x over a/Corollary 1

== Theorem ==
:$\map {\dfrac \d {\d x} } {\ln \size {x + \sqrt {x^2 + a^2} } } = \dfrac 1 {\sqrt {x^2 + a^2} }$
== Proof ==
We have that $\sqrt {x^2 + a^2} > x$ for all $x$.
Thus:
and the result follows.
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $14$.
Category:Derivative of Inverse Hyperbolic Sine
Derivative of Real Area Hyperbolic Sine of x over a/Corollary 2

== Theorem ==
:$\map {\dfrac \d {\d x} } {\ln \size {x - \sqrt {x^2 + a^2} } } = -\dfrac 1 {\sqrt {x^2 + a^2} }$
== Proof ==
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $14$.
Category:Derivative of Inverse Hyperbolic Sine
Derivative of Real Area Hyperbolic Tangent of x over a

== Theorem ==
:$\map {\dfrac \d {\d x} } {\map \artanh {\dfrac x a} } = \dfrac a {a^2 - x^2}$
where $-a < x < a$.
=== Corollary ===
== Proof ==
Let $-a < x < a$.
Then $-1 < \dfrac x a < 1$ and so:
$\artanh \dfrac x a$ is not defined when either $x \le -a$ or $x \ge a$.
== Also presented as ==
Some sources present this as:
:$\map {\dfrac \d {\d x} } {\dfrac 1 a \map \artanh {\dfrac x a} } = \dfrac 1 {a^2 - x^2}$
== Also see ==
* Derivative of $\arsinh \dfrac x a$
* Derivative of $\arcosh \dfrac x a$
* Derivative of $\arcoth \dfrac x a$
* Derivative of $\arsech \dfrac x a$
* Derivative of $\arcsch \dfrac x a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
Category:Derivative of Inverse Hyperbolic Tangent
Derivative of Real Area Hyperbolic Tangent of x over a/Corollary

== Theorem ==
:$\map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } } = \dfrac 1 {a^2 - x^2}$
where $\size x < a$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
Category:Derivative of Inverse Hyperbolic Tangent
Derivative of Reciprocal

== Theorem ==
Let $f: \R_{\ne 0} \to \R$ be the reciprocal function defined as:
:$\map f x = \dfrac 1 x$
Then its derivative is given by:
:$\map {f'} x = -\dfrac 1 {x^2}$
== Proof ==
We have:
== Sources ==
* : Appendix $2$: Table of derivatives and integrals of common functions
Category:Power Rule for Derivatives
Category:Reciprocals
Derivative of Riemann Zeta Function

== Theorem ==
The derivative of the Riemann zeta function is:
:$\ds \map {\zeta'} z = \frac {\d \zeta} {\d z} = -\sum_{n \mathop = 2}^\infty \frac {\map \ln n} {n^z}$
== Proof ==
== Also see ==
* Logarithmic Derivative of Riemann Zeta Function
Category:Derivatives
Category:Riemann Zeta Function
Derivative of Scalar 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 scalar triple product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} } = \dfrac {\d \mathbf a} {\d x} \cdot \paren {\mathbf b \times \mathbf c} + \mathbf a \cdot \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c} + \mathbf a \cdot \paren {\mathbf b \times \dfrac {\d \mathbf c} {\d x} }$
== Proof ==
== Also see ==
* Derivative of Dot Product of Vector-Valued Functions
* Derivative of Vector Cross Product of Vector-Valued Functions
* Derivative of Product of Real Function and Vector-Valued Function
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products
* : $\S 22$: Formulas involving Derivatives: $22.25$
Category:Differential Calculus
Category:Vector Calculus
Category:Scalar Triple Product
Derivative of Secant Function

== Theorem ==
:$\map {\dfrac \d {\d x} } {\sec x} = \sec x \tan x$
where $\cos x \ne 0$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Derivative of Sine Function
* Derivative of Cosine Function
* Derivative of Tangent Function
* Derivative of Cotangent Function
* Derivative of Cosecant Function
== Sources ==
* : $\text {II}$. Calculus: Differentiation
* : 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 Secant Function
Category:Derivatives of Trigonometric Functions
Category:Secant Function
Derivative of Secant Function/Proof 1

== Theorem ==
== Proof ==
From the definition of the secant function:
:$\sec x = \dfrac 1 {\cos x} = \paren {\cos x}^{-1}$
From Derivative of Cosine Function:
:$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
Then:
This is valid only when $\cos x \ne 0$.
Category:Derivative of Secant Function
Derivative of Secant of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\sec u} = \sec u \tan u \dfrac {\d u} {\d x}$
where $\sec$ is the secant function and $\tan$ is the tangent function.
== Proof ==
== Also see ==
* Derivative of Sine of Function
* Derivative of Cosine of Function
* Derivative of Tangent of Function
* Derivative of Cotangent of Function
* Derivative of Cosecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.18$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $7$
Category:Derivatives of Trigonometric Functions
Category:Secant Function
Derivative of Secant of a x

== Corollary to Derivative of Secant Function ==
:$\map {\dfrac \d {\d x} } {\sec a x} = a \sec a x \tan a x$
== Proof ==
== Also see ==
* Derivative of $\sin a x$
* Derivative of $\cos a x$
* Derivative of $\tan a x$
* Derivative of $\cot a x$
* Derivative of $\csc a x$
Category:Derivatives of Trigonometric Functions
Category:Secant Function
Derivative of Sequence of Holomorphic Functions

== Theorem ==
Let $U$ be an open, connected subset of $\C$.
Let $\sequence {f_n}$ be a sequence of holomorphic functions $f_n: U \to \C$.
Let $\sequence {f_n}$ converge pointwise to some function $f: U \to \C$.
Let $\sequence {f_n}$ converge uniformly on compact subsets of $U$.
Then $f$ is holomorphic on $U$.
Further, the sequence of derivatives $\sequence {f_n'}$ converges to $f'$ on $U$. 
This convergence is uniform on compact subsets of $U$.
== Proof ==
== Sources ==
* : $3$
Category:Holomorphic Functions