nyuuzyou/wikis huggingface dataset:
Derivative at Maximum or Minimum/Proof 1

== Theorem ==
== Proof ==
By definition of derivative at a point:
:$\dfrac {\map f x - \map f \xi} {x - \xi} \to \map {f'} \xi$ as $x \to \xi$
Suppose $\map {f'} \xi > 0$.
Then from Behaviour of Function Near Limit‎ it follows that:
:$\exists I = \openint {\xi - h} {\xi + h}: \dfrac {\map f x - \map f \xi} {x - \xi} > 0$
provided that $x \in I$ and $x \ne \xi$.
Now let $x_1$ be any number in the open interval $\openint {\xi - h} \xi$.
Then:
:$x_1 - \xi < 0$
and hence from:
:$\dfrac {\map f {x_1} - \map f \xi} {x_1 - \xi} > 0$
it follows that:
:$\map f {x_1} < \map f \xi$
Thus $f$ can not have a local minimum at $\xi$.
Now let $x_2$ be any number in the open interval $\openint \xi {\xi + h}$.
Then:
:$x_2 - \xi > 0$
and hence from:
:$\dfrac {\map f {x_2} - \map f \xi} {x_2 - \xi} > 0$
it follows that:
:$\map f {x_2} > \map f \xi$
Thus $f$ can not have a local maximum at $\xi$ either.
A similar argument can be applied to $-f$ to handle the case where $\map {f'} \xi < 0$.
The only other possibility is that $\map {f'} \xi = 0$, hence the result.
== Sources ==
* : $\S 11.2$
Category:Derivative at Maximum or Minimum
Derivative at Maximum or Minimum/Proof 2

== Theorem ==
== Proof ==
By definition of local maximum:
:$(1): \quad \map f {\xi + \epsilon} - \map f \xi < 0$
for sufficiently small $\epsilon \in \R_{>0}$.
Similarly, definition of local minimum:
:$(2): \quad \map f {\xi + \epsilon} - \map f \xi > 0$
for sufficiently small $\epsilon \in \R_{>0}$.
Let it be assumed that $\map f {\xi + \epsilon}$ can be expanded, using Taylor's Theorem, in positive integer powers of $\epsilon$
Then:
:$(3): \quad \map f {\xi + \epsilon} - \map f \xi = \epsilon \map {f'} \xi + \dfrac {\epsilon^2 \map {f' '} \xi} 2 + \map \OO {\epsilon^3}$
where $\map \OO {\epsilon^3}$ denotes big-$\OO$ notation.
That is, the quantity $\dfrac {\map \OO {\epsilon^3} } {\epsilon^3}$ is bounded.
From $(1)$ and $(2)$, at a local maximum or local minimum, the sign of $\map f {\xi + \epsilon} - \map f \xi$ is independent the sign of $\epsilon$ itself.
So from $(3)$ it follows that $\map {f'} \xi = 0$.
== Sources ==
* : Chapter $\text I$. The First Variation: $1.2$. Ordinary maximum and minimum theory
Category:Derivative at Maximum or Minimum
Derivative is Included in Closure

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then
:$A' \subseteq A^-$
where
:$A'$ denotes the derivative of $A$
:$A^-$ denotes the closure of $A$.
== Proof ==
Let $x \in A'$.
By Condition for Point being in Closure it is enough to prove that:
:for every open set $G$ of $T$:
:: if $x \in G$
:: then $A \cap G \ne \O$.
Let $G$ be an open set of $T$.
Let $x \in G$.
Then by Characterization of Derivative by Open Sets:
: there exists a point $\exists y \in S: y \in A \cap G \land x \ne y$.
 
Hence:
:$A \cap G \ne \O$
Hence the result.
== Sources ==
* 
* 
Category:Set Derivatives
Category:Set Closures
Derivative of Absolute Value Function

== Theorem ==
Let $\size x$ be the absolute value of $x$ for real $x$.
Then:
:$\dfrac \d {\d x} \size x = \dfrac x {\size x}$
for $x \ne 0$.
=== Corollary ===
== Proof ==
Now consider $x = 0$.
From the definition of derivative:
From Limit iff Limits from Left and Right, the limit does not exist.
== Also see ==
* Distributional Derivative of Absolute Value Function
Category:Derivative of Absolute Value Function
Category:Absolute Value Function
Category:Derivatives
Derivative of Absolute Value Function/Corollary

== Corollary to Derivative of Absolute Value Function ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\dfrac \d {\d x} \size u = \dfrac u {\size u} \dfrac {\d u} {\d x}$
for $u \ne 0$.
At $u = 0$, $\size u$ is not differentiable.
== Proof ==
From Derivative of Absolute Value Function:
:$\dfrac \d {\d u} \size u = \dfrac u {\size u}$
Hence by the Chain Rule for Derivatives:
:$\dfrac \d {\d x} \size u = \dfrac u {\size u} \dfrac {\d u} {\d x}$
Category:Derivatives
Category:Absolute Value Function
Derivative of Angular Component under Central Force

== Theorem ==
Let a point mass $p$ of mass $m$ be under the influence of a central force $\mathbf F$.
Let the position of $p$ at time $t$ be given in polar coordinates as $\polar {r, \theta}$.
Let $\mathbf r$ be the radius vector from the origin to $p$.
Then the rate of change of the angular coordinate of $p$ is inversely proportional to the square of the radial coordinate of $p$.
== Proof ==
Let $\mathbf F$ be expressed as:
:$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$
where:
:$\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$
:$\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$
:$F_r$ and $F_\theta$ are the magnitudes of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively.
From Motion of Particle in Polar Coordinates, the second order ordinary differential equations governing the motion of $m$ under the force $\mathbf F$ are:
However, we are given that $\mathbf F$ is a central force.
:
Thus:
for some constant $h$.
That is:
:$\dfrac {\d \theta} {\d t} = \dfrac h {r^2}$
Hence the result, by definition of inverse proportion.
== Sources ==
* : $\S 3.21$: Newton's Law of Gravitation: $(8)$
* : Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation
Category:Mechanics
Derivative of Arc Length

== Theorem ==
Let $C$ be a curve in the cartesian plane described by the equation $y = \map f x$.
Let $s$ be the length along the arc of the curve from some reference point $P$.
Then the derivative of $s$ with respect to $x$ is given by:
:$\dfrac {\d s} {\d x} = \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$
== Proof 1 ==
== Proof 2 ==
Category:Analytic Geometry
Category:Differential Calculus
Category:Integral Calculus
Category:Derivative of Arc Length
Derivative of Arc Length/Proof 1

== Theorem ==
== Proof ==
Consider a length $\d s$ of $C$, short enough for it to be approximated to a straight line segment:
:
By Pythagoras's Theorem, we have:
:$\d s^2 = \d x^2 + \d y^2$
Dividing by $\d x^2$ we have:
Hence the result, by taking the principal square root of both sides.
Category:Derivative of Arc Length
Derivative of Arccosecant Function

== Theorem ==
Let $x \in \R$ be a real number such that $\size x > 1$.
Let $\arccsc x$ denote the arccosecant of $x$.
Then:
:$\dfrac {\map \d {\arccsc x} } {\d x} = \dfrac {-1} {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\
\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\
\end{cases}$
=== Corollary ===
== Proof ==
Let $y = \arccsc x$ where $x < -1$ or $x > 1$.
Then:
Since $\dfrac {\d y} {\d x} = \dfrac {-1} {\csc y \cot y}$, the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\csc y \cot y$.
Writing $\csc y \cot y$ as $\dfrac {\cos y} {\sin^2 y}$, it is evident that the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\cos y$.
From Sine and Cosine are Periodic on Reals, $\cos y$ is never negative on its domain ($y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \land y \ne 0$).
However, by definition of the arccosecant of $x$:
:$0 < \arccsc x < \dfrac \pi 2 \implies x > 1$
:$-\dfrac \pi 2 < \arccsc x < 0 \implies x < -1$
Thus:
:$\dfrac {\map \d {\arccsc x} } {\d x} = \dfrac {-1} {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\
\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\
\end{cases}$
== Also see ==
* Derivative of Arcsine Function
* Derivative of Arccosine Function
* Derivative of Arctangent Function
* Derivative of Arccotangent Function
* Derivative of Arcsecant Function
== Sources ==
 
* 
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccosecant Function
Derivative of Arccosecant Function/Corollary

== Corollary to Derivative of Arccosecant Function ==
Let $x \in \R$.
Let $\arccsc \dfrac x a$ be the arccosecant of $\dfrac x a$.
Then:
:$\map {\dfrac \d {\d x} } {\map \arccsc {\dfrac x a} } = \dfrac {-a} {\size x {\sqrt {x^2 - a^2} } } = \begin{cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end{cases}$
== Proof ==
Similarly:
== Also see ==
* Derivative of $\arcsin \dfrac x a$
* Derivative of $\arccos \dfrac x a$
* Derivative of $\arctan \dfrac x a$
* Derivative of $\arccot \dfrac x a$
* Derivative of $\arcsec \dfrac x a$
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccosecant Function
Derivative of Arccosecant of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$ such that $\size u > 1$.
Then:
:$\map {\dfrac \d {\d x} } {\arccsc u} = -\dfrac 1 {\size u \sqrt {u^2 - 1} } \dfrac {\d u} {\d x}$
where $\arcsec$ denotes the arccosecant of $x$.
== Proof ==
== Also see ==
* Derivative of Arcsine of Function
* Derivative of Arccosine of Function
* Derivative of Arctangent of Function
* Derivative of Arccotangent of Function
* Derivative of Arcsecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.25$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $18$
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccosecant Function
Derivative of Arccosine Function

== Theorem ==
Let $x \in \R$ be a real number such that $-1 < x < 1$.
Let $\arccos x$ be the arccosine of $x$.
Then:
:$\map {D_x} {\arccos x} = \dfrac {-1} {\sqrt {1 - x^2} }$
=== Corollary ===
== Proof ==
Let $y = \arccos x$ where $-1 < x < 1$.
Then:
:$x = \cos y$
Then from Derivative of Cosine Function:
:$\dfrac {\d x} {\d y} = -\sin y$
Hence from Derivative of Inverse Function:
:$\dfrac {\d y} {\d x} = \dfrac {-1} {\sin y}$
From Sum of Squares of Sine and Cosine, we have:
:$\cos^2 y + \sin^2 y = 1 \implies \sin y = \pm \sqrt {1 - \cos^2 y}$
Now $\sin y \ge 0$ on the range of $\arccos x$, that is, for $y \in \closedint 0 \pi$.
Thus it follows that we need to take the positive root of $\sqrt {1 - \cos^2 y}$.
So:
:$\dfrac {\d y} {\d x} = \dfrac {-1} {\sqrt {1 - \cos^2 y} }$
and hence the result.
== Also see ==
* Derivative of Arcsine Function
* Derivative of Arctangent Function
* Derivative of Arccotangent Function
* Derivative of Arcsecant Function
* Derivative of Arccosecant Function
== Sources ==
* : $\S 16.5 \ (3)$
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $4.$ Inverse trigonometric functions
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccosine Function
Derivative of Arccosine Function/Corollary

== Corollary to Derivative of Arccosine Function ==
Let $a \in \R$ be a constant
Let $x \in \R$ be a real number such that $x^2 < a^2$.
Let $\map \arccos {\dfrac x a}$ be the arccosine of $\dfrac x a$.
Then:
:$\map {\dfrac \d {\d x} } {\map \arccos {\dfrac x a} } = \dfrac {-1} {\sqrt {a^2 - x^2} }$
== Proof ==
== Also see ==
* Derivative of $\arcsin \dfrac x a$
* Derivative of $\arctan \dfrac x a$
* Derivative of $\arccot \dfrac x a$
* Derivative of $\arcsec \dfrac x a$
* Derivative of $\arccsc \dfrac x a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $12$.
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccosine Function
Derivative of Arccosine of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$ such that $\size {\map u x} < 1$, that is, $0 < \arccos u < \pi$.
Then:
:$\map {\dfrac \d {\d x} } {\arccos u} = -\dfrac 1 {\sqrt {1 - u^2} } \dfrac {\d u} {\d x}$
where $\arccos$ denotes the arccosine of $x$.
== Proof ==
== Also see ==
* Derivative of Arcsine of Function
* Derivative of Arctangent of Function
* Derivative of Arccotangent of Function
* Derivative of Arcsecant of Function
* Derivative of Arccosecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.21$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $14$
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccosine Function
Derivative of Arccotangent Function

== Theorem ==
Let $x \in \R$.
Let $\arccot x$ be the arccotangent of $x$.
Then:
:$\dfrac {\map \d {\arccot x} } {\d x} = \dfrac {-1} {1 + x^2}$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also see ==
* Derivative of Arcsine Function
* Derivative of Arccosine Function
* Derivative of Arctangent Function
* Derivative of Arcsecant Function
* Derivative of Arccosecant Function
== Sources ==
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $4.$ Inverse trigonometric functions
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccotangent Function
Category:Derivative of Arccotangent Function
Derivative of Arccotangent Function/Also presented as

== Derivative of Arccotangent Function: Also presented as == 
The derivative of the crccotangent function can also be presented in the form:
:$\dfrac {\map \d {\arccot x} } {\d x} = \dfrac {-1} {x^2 + 1}$
Category:Derivative of Arccotangent Function
Derivative of Arccotangent Function/Corollary

== Corollary to Derivative of Arccotangent Function ==
Let $x \in \R$.
Let $\arccot \dfrac x a$ be the arccotangent of $\dfrac x a$.
Then:
:$\map {\dfrac \d {\d x} } {\map \arccot {\dfrac x a} } = \dfrac {-a} {a^2 + x^2}$
== Proof ==
== Also defined as ==
This result can also be reported as:
:$\map {\dfrac \d {\d x} } {\map \arccot {\dfrac x a} } = \dfrac {-a} {x^2 + a^2}$
== Also see ==
* Derivative of $\arcsin \dfrac x a$
* Derivative of $\arccos \dfrac x a$
* Derivative of $\arctan \dfrac x a$
* Derivative of $\arcsec \dfrac x a$
* Derivative of $\arccsc \dfrac x a$
Category:Derivative of Arccotangent Function
Derivative of Arccotangent of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\arccot u} = -\dfrac 1 {1 + u^2} \dfrac {\d u} {\d x}$
where $\arccot$ denotes the arccotangent of $x$.
== Proof ==
== Also see ==
* Derivative of Arcsine of Function
* Derivative of Arccosine of Function
* Derivative of Arctangent of Function
* Derivative of Arcsecant of Function
* Derivative of Arccosecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.23$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $16$
Category:Derivatives of Inverse Trigonometric Functions
Category:Arccotangent Function
Derivative of Arcsecant Function

== Theorem ==
Let $x \in \R$ be a real number such that $\size x > 1$.
Let $\arcsec x$ be the arcsecant of $x$.
Then:
:$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {+1} {x \sqrt {x^2 - 1} } & : 0 < \arcsec x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\
\dfrac {-1} {x \sqrt {x^2 - 1} } & : \dfrac \pi 2 < \arcsec x < \pi \ (\text {that is: $x < -1$}) \\
\end{cases}$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $y = \arcsec x$ where $\size x > 1$.
Then:
Since $\dfrac {\d y} {\d x} = \dfrac 1 {\sec y \tan y}$, the sign of $\dfrac {\d y} {\d x}$ is the same as the sign of $\sec y \tan y$.
Writing $\sec y \tan y$ as $\dfrac {\sin y} {\cos^2 y}$, it is evident that the sign of $\dfrac {\d y} {\d x}$ is the same as the sign of $\sin y$.
From Sine and Cosine are Periodic on Reals, $\sin y$ is never negative on its domain ($y \in \closedint 0 \pi \land y \ne \pi/2$).
However, by definition of the arcsecant of $x$:
:$0 < \arcsec x < \dfrac \pi 2 \implies x > 1$
:$\dfrac \pi 2 < \arcsec x < \pi \implies x < -1$
Thus:
:$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {\size x \sqrt {x^2 - 1} } = \begin{cases} \dfrac {+1} {x \sqrt {x^2 - 1} } & : 0 < \arcsec x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\
\dfrac {-1} {x \sqrt {x^2 - 1} } & : \dfrac \pi 2 < \arcsec x < \pi \ (\text {that is: $x < -1$}) \\
\end{cases}$
Hence the result.
== Also see ==
* Derivative of Arcsine Function
* Derivative of Arccosine Function
* Derivative of Arctangent Function
* Derivative of Arccotangent Function
* Derivative of Arccosecant Function
== Sources ==
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
Category:Derivatives of Inverse Trigonometric Functions
Category:Arcsecant Function
Category:Derivative of Arcsecant Function
Derivative of Arcsecant Function/Corollary 1

== Corollary to Derivative of Arcsecant Function ==
Let $x \in \R$.
Let $\arcsec \dfrac x a$ be the arcsecant of $\dfrac x a$.
Then:
:$\map {\dfrac \d {\d x} } {\map \arcsec {\dfrac x a} } = \dfrac a {\size x \sqrt {x^2 - a^2} } = \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \ (\text {that is: $x > a$}) \\
\dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \ (\text {that is: $x < -a$}) \\
\end{cases}$
== Proof ==
Similarly:
== Also see ==
* Derivative of $\arcsin \dfrac x a$
* Derivative of $\arccos \dfrac x a$
* Derivative of $\arctan \dfrac x a$
* Derivative of $\arccot \dfrac x a$
* Derivative of $\arccsc \dfrac x a$
Category:Derivative of Arcsecant Function
Derivative of Arcsecant Function/Corollary 2

== Theorem ==
Let $x \in \R$ be a real number such that $x < -1$ or $x > 1$.
Let $\arcsec x$ be the arcsecant of $x$.
Then:
:$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {x^2 \sqrt {1 - \frac 1 {x^2} } }$
== Proof ==
From Derivative of Arcsecant Function:
:$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {\size x \sqrt {x^2 - 1} }$
Since for all $x \in \R$, we have $\size x = \sqrt{x^2}$, we can write:
:$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {\sqrt {x^2} \sqrt {x^2 - 1} }$
Multiplying the denominator by $1 = \dfrac {\sqrt{x^2} } {\sqrt{x^2} }$ yields:
:$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {x^2 \sqrt {1 - \frac 1 {x^2} } }$
== Sources ==
* 
Category:Derivative of Arcsecant Function
Derivative of Arcsecant of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$ such that $\size u > 1$.
Then:
:$\map {\dfrac \d {\d x} } {\arcsec u} = \dfrac 1 {\size u \sqrt {u^2 - 1} } \dfrac {\d u} {\d x}$
where $\arcsec$ denotes the arcsecant of $x$.
== Proof ==
== Also see ==
* Derivative of Arcsine of Function
* Derivative of Arccosine of Function
* Derivative of Arctangent of Function
* Derivative of Arccotangent of Function
* Derivative of Arccosecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.24$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $17$
Category:Derivatives of Inverse Trigonometric Functions
Category:Arcsecant Function
Derivative of Arcsine Function

== Theorem ==
Let $x \in \R$ be a real number such that $\size x < 1$, that is, $\size {\arcsin x} < \dfrac \pi 2$.
Let $\arcsin x$ be the real arcsine of $x$.
Then:
:$\dfrac {\map \d {\arcsin x} } {\d x} = \dfrac 1 {\sqrt {1 - x^2} }$
=== Corollary ===
== Proof ==
== Also see ==
* Derivative of Arccosine Function
* Derivative of Arctangent Function
* Derivative of Arccotangent Function
* Derivative of Arcsecant Function
* Derivative of Arccosecant Function
== Sources ==
* : $\text {II}$. Calculus: Differentiation
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $4.$ Inverse trigonometric functions
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
Category:Derivative of Arcsine Function
Category:Derivatives of Inverse Trigonometric Functions
Category:Arcsine Function
Derivative of Arcsine Function/Corollary

== Corollary to Derivative of Arcsine Function ==
Let $a \in \R$ be a constant
Let $x \in \R$ be a real number such that $x^2 < a^2$.
Let $\map \arcsin {\dfrac x a}$ be the arcsine of $\dfrac x a$.
Then:
:$\map {\dfrac \d {\d x} } {\map \arcsin {\dfrac x a} } = \dfrac 1 {\sqrt {a^2 - x^2} }$
== Proof ==
== Also see ==
* Derivative of $\arccos \dfrac x a$
* Derivative of $\arctan \dfrac x a$
* Derivative of $\arccot \dfrac x a$
* Derivative of $\arcsec \dfrac x a$
* Derivative of $\arccsc \dfrac x a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $12$.
Category:Derivatives of Inverse Trigonometric Functions
Category:Arcsine Function
Derivative of Arcsine Function/Proof

== Theorem ==
== Proof ==
Let $y = \arcsin x$ where $-1 < x < 1$.
Then:
Then:
Now $\cos y \ge 0$ on the image of $\arcsin x$, that is:
:$y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
Thus it follows that we need to take the positive root of $\sqrt {1 - \sin^2 y}$.
So:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Inverse Ratios
* : $\S 16.5 \ (3)$
Category:Derivative of Arcsine Function
Derivative of Arcsine of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$ such that $\size {\map u x} < 1$, that is, $\size {\arcsin u} < \dfrac \pi 2$.
Then:
:$\map {\dfrac \d {\d x} } {\arcsin u} = \dfrac 1 {\sqrt {1 - u^2} } \dfrac {\d u} {\d x}$
where $\arcsin$ denotes the real arcsine of $x$.
== Proof ==
== Also see ==
* Derivative of Arccosine of Function
* Derivative of Arctangent of Function
* Derivative of Arccotangent of Function
* Derivative of Arcsecant of Function
* Derivative of Arccosecant of Function
== Sources ==
* : $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.20$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $13$
Category:Derivatives of Inverse Trigonometric Functions
Category:Arcsine Function
Derivative of Arctangent Function

== Theorem ==
Let $x \in \R$.
Let $\arctan x$ be the arctangent of $x$.
Then:
:$\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {1 + x^2}$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also see ==
* Derivative of Arcsine Function
* Derivative of Arccosine Function
* Derivative of Arccotangent Function
* Derivative of Arcsecant Function
* Derivative of Arccosecant Function
== Sources ==
* : $\text {II}$. Calculus: Differentiation
* : $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $4.$ Inverse trigonometric functions
* : $\S 16.5 \ (4)$
* 
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions
* : Appendix: Table $1$: Derivatives
* : Appendix: Table $1$: Derivatives
* 
* : Appendix $6$: Derivatives
* : Appendix $7$: Derivatives
Category:Derivative of Arctangent Function
Category:Arctangent Function
Category:Derivatives of Inverse Trigonometric Functions
Derivative of Arctangent Function/Corollary

== Corollary to Derivative of Arctangent Function ==
Let $x \in \R$.
Let $\map \arctan {\dfrac x a}$ denote the arctangent of $\dfrac x a$.
Then:
:$\map {\dfrac \d {\d x} } {\map \arctan {\dfrac x a} } = \dfrac a {a^2 + x^2}$
== Proof ==
== Also presented as ==
This result can also be presented as:
:$\map {\dfrac \d {\d x} } {\map \arctan {\dfrac x a} } = \dfrac a {x^2 + a^2}$
== Also see ==
* Derivative of $\arcsin \dfrac x a$
* Derivative of $\arccos \dfrac x a$
* Derivative of $\arccot \dfrac x a$
* Derivative of $\arcsec \dfrac x a$
* Derivative of $\arccsc \dfrac x a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $13$.
Category:Derivative of Arctangent Function