nyuuzyou/wikis huggingface dataset:
Derivative of Arctangent of Function

== Theorem ==
Let $u$ be a differentiable real function of $x$.
Then:
:$\map {\dfrac \d {\d x} } {\arctan u} = \dfrac 1 {1 + u^2} \dfrac {\d u} {\d x}$
where $\arctan$ denotes the arctangent of $x$.
== Proof ==
== Also see ==
* Derivative of Arcsine of Function
* Derivative of Arccosine 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: $13.22$
* : Chapter $4$. Derivatives: Derivatives of Special Functions: $15$
Category:Derivatives of Inverse Trigonometric Functions
Category:Arctangent Function
Derivative of Complex Composite Function

== Theorem ==
Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is an open set.
Let $g: \Img f \to \C$ be a complex-differentiable function, where $\Img f$ denotes the image of $f$.
Define $h = f \circ g: D \to C$ as the composite of $f$ and $g$.
Then $h$ is complex-differentiable, and its derivative is defined as:
:$\forall z \in D: \map {h'} z = \map {f'} {\map g z} \map {g'} z$
== Proof ==
Put $y = \map g z$.
Let $\delta z \in \C \setminus \set 0$.
Put $\delta y = \map g {z + \delta z} - y$, so:
:$\map g {z + \delta z} = y + \delta y$
As $\delta z \to 0$, we have:
:$(1): \quad \delta y \to 0$ by definition of continuity, as $g$ is continuous.
:$(2): \quad \dfrac {\delta y} {\delta z} \to \map {g'} z$ by definition of complex-differentiability.
There are two cases to consider:
=== Case 1 ===
Suppose $\map {g'} z \ne 0$.
When $\delta z \to 0$, we have $\delta y \ne 0$ from $(2)$, if $\cmod {\delta z}$ is sufficiently small.
Then:
hence the result.
=== Case 2 ===
Now suppose $\map {g'} z = 0$.
When $\delta z \to 0$, there are two possibilities:
=== Case 2a ===
If $\delta y = 0$, then:
:$\dfrac {\map h {z + \delta z} - \map h z} {\delta z} = 0 = \map {f'} y \map {g'} z$
Hence the result.
=== Case 2b ===
If $\delta y \ne 0$, then:
:$\dfrac {\map h {z + \delta z} - \map h z} {\delta z} = \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \dfrac {\delta y} {\delta z}$
As $\delta y \to 0$:
:$(1): \quad \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \to \map {f'} y$
:$(2): \quad \dfrac {\delta y} {\delta z} \to 0$
Thus:
:$\ds \lim_{\delta z \mathop \to 0} \frac {\map h {z + \delta z} - \map h z} {\delta z} \to 0 = \map {f'} y \map {g'} z$
Again, hence the result.
All cases have been covered, so by Proof by Cases, the result is complete.
== Also known as ==
This is often informally referred to as the chain rule (for differentiation).
== Also see ==
* Derivative of Composite Function.
== Sources ==
* : $\S 1.1$
Category:Complex Differential Calculus
Derivative of Complex Polynomial

== Theorem ==
Let $a_n \in \C$ for $n \in \set {0, 1, \ldots, N}$, where $N \in \N$.
Let $f: \C \to \C$ be a complex polynomial defined by $\ds \map f z = \sum_{n \mathop = 0}^N a_n z^n$.
Then $f$ is complex differentiable and its derivative is:
:$\ds \map {f'} z = \sum_{n \mathop = 1}^N n a_n z^{n - 1}$
== Proof ==
For all $n > N$, put $a_n = 0$.
Then:
:$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$
The result now follows from Derivative of Complex Power Series.
Category:Complex Differential Calculus
Derivative of Complex Power Series

== Theorem ==
Let $\xi \in \C$ be a complex number.
Let $\sequence {a_n}$ be a sequence in $\C$.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a power series in a complex variable $z \in \C$ about $\xi$.
Let $R$ be the radius of convergence of the series defining $\map f z$.
Let $\cmod {z - \xi} < R$.
Then $f$ is complex-differentiable and its derivative is:
:$\ds \map {f'} z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
== Proof 1 ==
== Proof 2 ==
Category:Derivative of Complex Power Series
Category:Complex Differential Calculus
Category:Complex Power Series
Category:Complex Analysis
Derivative of Complex Power Series/Proof 1

== Theorem ==
== Proof ==
Define:
:$\ds \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
From Radius of Convergence of Derivative of Complex Power Series, it follows that $g$ has radius of convergence $R$. 
Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$.
Define:
:$\ds M = \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^n$
Suppose that $\cmod h \le R - \epsilon - \cmod {z - \xi}$.
Then, by the Binomial Theorem and the Triangle Inequality:
Letting $h \to 0$ we see that $\map {f'} z$ exists and:
:$\map {f'} z = \map g z$
as desired.
== Also see ==
* The proof for real power series is identical.
Category:Derivative of Complex Power Series
Derivative of Complex Power Series/Proof 2

== Theorem ==
== Proof ==
=== Lemma ===
Define:
:$\ds \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$.
Let:
:$\ds M = \sum_{n \mathop = 2}^\infty \dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2}$
We use the Root Test to prove convergence of this series:
The last equality follows from the lemma and:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$
Suppose that $\size h \le R - \epsilon - \cmod {z - \xi}$.
It follows by the Triangle Inequality that:
:$\cmod {z - \xi + h} \le \cmod {z - \xi} + \size h \le R - \epsilon$
By the Triangle Inequality, Difference of Two Powers, and Closed Form for Triangular Numbers, the following holds:
Letting $h \to 0$ we see that $\map {f'} z$ exists and $\map {f'} z = \map g z$, as desired.
== Also see ==
The proof for real power series is identical.
Category:Derivative of Complex Power Series
Derivative of Complex Power Series/Proof 2/Lemma

== Lemma ==
Let $n \in N_{\ge 1}$.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$
== Proof ==
Choose any $\alpha > 1$. 
It follows from the ratio test that:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 {\alpha^n} \frac {n \paren {n - 1} } 2 = 0$
Therefore, for all sufficiently large $n$:
:$\dfrac {n \paren {n - 1} } 2 \le \alpha^n$
and so:
:$\paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le \alpha$
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le \alpha$
Since $\alpha > 1$ was arbitrary, we can conclude that:
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le 1$
It is clear that the following holds for sufficiently large $n$:
:$\ds \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \ge 1^{1/n} = 1$
Therefore:
:$\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$
Category:Derivative of Complex Power Series
Derivative of Composite Function

== Theorem ==
Let $I, J$ be open real intervals.
Let $g : I \to J$ and $f : J \to \R$ be real functions.
Let $h : I \to \R$ be the real function defined as:
:$\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for each $x_0 \in I$ such that:
:$g$ is differentiable at $x_0$
:$f$ is differentiable at $\map g {x_0}$
it holds that $h$ is differentiable at $x_0$ and:
:$\map {h'} {x_0} = \map {f'} {\map g {x_0}} \map {g'} {x_0}$
where $h'$ denotes the derivative of $h$.
Using the $D_x$ notation:
:$\map {D_x} {\map f {\map g x} } = \map {D_{\map g x} } {\map f {\map g x} } \map {D_x} {\map g x}$
=== Corollary ===
=== Second Derivative ===
=== Third Derivative ===
=== $3$ Functions ===
=== Jacobians ===
== Also presented as ==
== Also known as ==
== Proof 1 ==
== Proof 2 ==
== Informal Proof ==
== Examples ==
== Also see ==
* Chain Rule for Real-Valued Functions
* Faà di Bruno's Formula
== Sources ==
* : $1$ Partial Differentiation: $\S 3$. Functions of Several Variables
* 
* 
* 
* 
* : Chapter $5$: Differentiation of Functions of One Real Variable: $\text {5-4}$ The chain rule
* : $\S 10.13$
Category:Derivative of Composite Function
Category:Differential Calculus
Derivative of Composite Function/3 Functions

== Theorem ==
Let $f, g, h$ be continuous real functions such that:
Then:
:$\dfrac {\d y} {\d x} = \dfrac {\d y} {\d u} \cdot \dfrac {\d u} {\d v} \cdot \dfrac {\d v} {\d x}$
== Proof ==
== Sources ==
* 
* 
Category:Derivative of Composite Function
Derivative of Composite Function/Also known as

== Derivative of Composite Function: Also known as == 
The rule to calculate the Derivative of Composite Function is often informally referred to as:
:the Chain Rule for Derivatives
:the Chain Rule for Differentiation.
Some sources refer to it as just the Chain Rule, but the latter term has a number of applications.
== Sources ==
* : Chapter $5$: Differentiation of Functions of One Real Variable: $\text {5-4}$ The chain rule
* : $\S 13$: General Rules of Differentiation: $13.11$
* : $5$. Differential Calculus: Appendix: Differentiation Rules: $5.$ Chain rule
* : $\S 1.2.5$: Permutations and Factorials: Exercise $21$
* 
* 
* 
* 
* 
* 
Category:Derivative of Composite Function
Derivative of Composite Function/Also presented as

== Derivative of Composite Function: Also presented as ==
The Derivative of Composite Function is also often seen presented using Leibniz's notation for derivatives: 
:$\dfrac {\d y} {\d x} = \dfrac {\d y} {\d u} \cdot \dfrac {\d u} {\d x}$
or:
:$\dfrac \d {\d x} \map u v = \dfrac {\d u} {\d v} \cdot \dfrac {\d v} {\d x}$
where $\dfrac {\d y} {\d x}$ denotes the derivative of $y$ $x$.
Some sources go so far as to mix their notation and present something like this:
:$y' = \dfrac {\d f} {\d g} \map {g'} x$
Also to be mentioned is:
:$D_x^1 w = D_u^1 w \, D_x^1 u$
or:
:${D_x}^1 w = {D_u}^1 w \, {D_x}^1 u$
where ${D_x}^k u$ denotes the $k$th derivative of $u$ $x$.
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Double Function
* : $1$ Partial Differentiation: $\S 3$. Functions of Several Variables
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Derivatives: $3.3.5$
* : $\S 13$: General Rules of Differentiation: $13.11$
* : $5$. Differential Calculus: Appendix: Differentiation Rules: $5.$ Chain rule
* : $\S 10.13$
* : $\S 1.2.5$: Permutations and Factorials: Exercise $21$
* 
* 
Category:Derivative of Composite Function
Derivative of Composite Function/Corollary

== Theorem ==
Let $f, g, h$ be continuous real functions such that:
:$y = \map f u, x = \map g u$
Then:
:$\dfrac {\d y} {\d x} = \dfrac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x} {\d u} } }$
for $\dfrac {\d x} {\d u} \ne 0$.
== Proof ==
== Sources ==
* : $\S 13$: General Rules of Differentiation: $13.13$
Category:Derivative of Composite Function
Derivative of Composite Function/Examples

== Examples of Derivatives of Composite Functions ==
=== Example: $\paren {3 x + 1}^2$ ===
=== Example: $\map \sin {x^2}$ ===
=== Example: $\sqrt {1 + x}$ ===
=== Example: $\sqrt {\sin x}$ ===
=== Example: $\paren {3 x + 1}^4$ ===
=== Example: $\dfrac 1 {\paren {2 x + 1}^3}$ ===
=== Example: $\sqrt {x^2 + x + 1}$ ===
=== Example: $\sin^3 2 x$ ===
=== Example: $\map {\cos^2} {a x + b}$ ===
=== Example: $\sqrt {\arcsin x}$ ===
=== Example: $\dfrac 1 {\arctan x}$ ===
=== Example: $e^{a x^2}$ ===
=== Example: $\map \ln {1 + x^3}$ ===
=== Example: $\ln \cosec x$ ===
=== Example: $\sqrt {\dfrac {x + 1} {x - 1} }$ ===
=== Example: $\ln \dfrac x {x + 1}$ ===
=== Example: $e^{3 x^2 + 4}$ ===
=== Example: $e^{x^2 + x + 1}$ ===
=== Example: $\sqrt {x^2 + 1}$ ===
=== Example: $\paren {a x^2 + b x + c}^3$ ===
=== Example: $a^{\sin x}$ ===
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/(3x+1)^2

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\paren {3 x + 1}^2} = 6 \paren {3 x + 1}$
== Proof ==
Let $u = 3 x + 1$.
Let $y = u^2$.
Then we have:
:$y = \paren {3 x + 1}^2$
and so:
== Sources ==
* 
* 
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/(3x+1)^4

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\paren {3 x + 1}^4} = 12 \paren {3 x + 1}^3$
== Proof ==
Let $u = 3 x + 1$.
Let $y = u^4$.
Then we have:
:$y = \paren {3 x + 1}^4$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $3$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/(a x^2 + b x + c)^3

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\paren {a x^2 + b x + c}^3} = 3 \paren {2 a x + b} \paren {a x^2 + b x + c}^2$
== Proof ==
Let $u = a x^2 + b x + c$.
Let $y = u^3$.
Thus we have:
:$y = \paren {a x^2 + b x + c}^3$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $29$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Exponential of 3 x^2 + 4

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {e^{3 x^2 + 4} } = 6 x e^{3 x^2 + 4}$
== Proof ==
Let $u = 3 x^2 + 4$.
Let $y = e^u$.
Thus we have:
:$y = e^{3 x^2 + 4}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $23$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Exponential of a x^2

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {e^{a x^2} } = 2 a x e^{a x^2}$
== Proof ==
Let $u = a x^2$.
Let $y = e^u$.
Thus we have:
:$y = e^{a x^2}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $13$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Exponential of x^2 + x + 1

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {e^{x^2 + x + 1} } = \paren {2 x + 1} e^{x^2 + x + 1}$
== Proof ==
Let $u = x^2 + x + 1$.
Let $y = e^u$.
Thus we have:
:$y = e^{x^2 + x + 1}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $24$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Logarithm of 1 + x^3

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\map \ln {1 + x^3} } = \dfrac {3 x^2} {1 + x^3}$
== Proof ==
Let $u = 1 + x^3$.
Let $y = \ln u$.
Thus we have:
:$y = \map \ln {1 + x^3}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $14$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Logarithm of Cosecant of x

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\ln \cosec x} = -\cot x$
== Proof ==
Let $u = \cosec x$.
Let $y = \ln u$.
Thus we have:
:$y = \ln \cosec x$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $15$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Logarithm of x over x + 1

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\ln \dfrac x {x + 1} } = \dfrac 1 {x \paren {x + 1} }$
== Proof ==
Let $u = \dfrac x {x + 1}$.
Let $y = \ln u$.
Thus we have:
:$y = \ln \dfrac x {x + 1}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $19$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Reciprocal of (2x+1)^3

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\dfrac 1 {\paren {2 x + 1}^3} } = -\dfrac 6 {\paren {2 x + 1}^4}$
== Proof ==
Let $u = 2 x + 1$.
Let $y = u^{-3}$.
Then we have:
:$y = \paren {2 x + 1}^{-3}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $4$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Reciprocal of Arctangent of x

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\dfrac 1 {\arctan x} } = -\dfrac 1 {\paren {1 + x^2} \arctan^2 x}$
== Proof ==
Let $u = \arctan x$.
Let $y = u^{-1}$.
Thus we have:
:$y = \dfrac 1 {\arctan x}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $12$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Root of 1 + x

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\sqrt {1 + x} } = \dfrac 1 {2 \sqrt {1 + x} }$
== Proof ==
Let $u = 1 + x$.
Let $y = u^{1/2}$.
Thus by definition of square root we have:
:$y = \paren {1 + x}^{1/2}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $1$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Root of Arcsine of x

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\sqrt {\arcsin x} } = \dfrac 1 {2 \sqrt {\paren {1 - x^2} \arcsin x } }$
== Proof ==
Let $u = \arcsin x$.
Let $y = u^{1/2}$.
Thus we have:
:$y = \paren {\arcsin x}^{1/2}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $11$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Root of sin x

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\sqrt {\sin x} } = \dfrac {\cos x} {2 \sqrt {\sin x} }$
== Proof ==
Let $u = \sin x$.
Let $y = u^{1/2}$.
Thus by definition of square root we have:
:$y = \paren {\sin x}^{1/2}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $2$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Root of x^2 + 1

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\sqrt {x^2 + 1} } = \dfrac x {\sqrt {x^2 + 1} }$
== Proof ==
Let $u = x^2 + 1$.
Let $y = u^{1/2}$.
Thus we have:
:$y = \sqrt {x^2 + 1}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $27$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Root of x^2 + x + 1

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\sqrt {x^2 + x + 1} } = \dfrac {2 x + 1} {2 \sqrt {x^2 + x + 1} }$
== Proof ==
Let $u = x^2 + x + 1$.
Let $y = u^{1/2}$.
Then we have:
:$y = \paren {x^2 + x + 1}^{1/2}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $5$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Root of x + 1 over x - 1

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\sqrt {\dfrac {x + 1} {x - 1} } } = -\dfrac 1 {\sqrt {\paren {x + 1} \paren {x - 1}^3} }$
== Proof ==
Let $u = \dfrac {x + 1} {x - 1}$.
Let $y = u^{1/2}$.
Thus we have:
:$y = \paren {\dfrac {x + 1} {x - 1} }^{1/2}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $18$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Sine of 2 x Cubed

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\sin^3 2 x} = 6 \sin^2 2 x \cos 2 x$
== Proof ==
Let $u = \sin 2 x$.
Let $y = u^3$.
Thus we have:
:$y = \sin^3 2 x$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $8$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/Square of Cosine of a x + b

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\map {\cos^2} {a x + b} } = -2 a \map \cos {a x + b} \map \sin {a x + b}$
== Proof ==
Let $u = \map \cos {a x + b}$.
Let $y = u^2$.
Thus we have:
:$y = \map {\cos^2} {a x + b}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $10$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/a^sin x

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {a^{\sin x} } = \cos x a^{\sin x} \ln a$
== Proof ==
Let $u = \sin x$.
Let $y = a^u$.
Thus we have:
:$y = a^{\sin x}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $30$.
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Examples/sin(x^2)

== Example of Derivative of Composite Function ==
:$\map {\dfrac \d {\d x} } {\map \sin {x^2} } = 2 x \map \cos {x^2}$
== Proof ==
Let $y = x^2$.
Let $z = \sin y$.
Then we have:
:$z = \map \sin {x^2}$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Double Function
Category:Examples of Derivatives of Composite Functions
Derivative of Composite Function/Informal Proof

== Theorem ==
== Proof ==
Some sources, introducing the Derivative of Composite Function at elementary level, provide the following non-rigorous argument:
:If $z$ is a function of $y$ where $y$ itself is some function of $x$,
::it is obvious that:
:::$\dfrac {\delta z} {\delta x} = \dfrac {\delta z} {\delta y} \cdot \dfrac {\delta y} {\delta x}$
::since these quantities being finite can be cancelled as in Arithmetic.
:If we now let the quantities concerned tend to zero, taking the limit, we get
:::$\dfrac {\d z} {\d x} = \dfrac {\d z} {\d y} \cdot \dfrac {\d y} {\d x}$
However, this informal argument is insufficiently rigorous for .
Hence, this must not be interpreted to mean that derivatives can be treated as fractions.
It simply is a convenient notation.
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Double Function
Category:Derivative of Composite Function
Derivative of Composite Function/Proof 1

== Theorem ==
== Proof ==
Let $f, g, x_0$ satisfy the conditions of the theorem.
Define $g^* : I \to \R$ as:
:$\map {g^*} x = \begin{cases} \dfrac {\map g x - \map g {x_0}} {x - x_0} & : x \ne x_0 \\ \map {g'} {x_0} & : x = x_0 \end{cases}$
Then, for every $x \in I$:
:$\paren 1: \quad \map g x - \map g {x_0} = \paren {x - x_0} \map {g^*} x$
for when $x = x_0$, both sides are equal to $0$.
Furthermore, by the definition of derivative, it follows that $g^*$ is continuous at $x_0$.
Define $y_0 := \map g {x_0}$
In the same way, define $f^* : J \to \R$ as:
:$\map {f^*} y = \begin{cases} \dfrac {\map f y - \map f {y_0}} {y - y_0} & : y \ne y_0 \\ \map {f'} {y_0} & : y = y_0 \end{cases}$
Likewise, for every $y \in J$:
:$\paren 2: \quad \map f y - \map f {y_0} = \paren {y - y_0} \map {f^*} y$
and $f^*$ is continuous at $y_0$.
Now, for every $x \in I$, we have:
Therefore, we have:
Therefore, by definition of derivative:
:$\map {h'} {x_0} = \map {f'} {\map g {x_0}} \map {g'} {x_0}$
== Also presented as ==
Category:Derivative of Composite Function
Derivative of Composite Function/Proof 2

== Theorem ==
== Proof ==
Let $\map g x = y$, and let:
Thus:
:$\delta y \to 0$ as $\delta x \to 0$
and:
:$(1): \quad \dfrac {\delta y} {\delta x} \to \map {g'} x$
There are two cases to consider:
=== Case 1 ===
Suppose $\map {g'} x \ne 0$ and that $\delta x$ is small but non-zero.
Then $\delta y \ne 0$ from $(1)$ above, and:
hence the result.
=== Case 2 ===
Now suppose $\map {g'} x = 0$ and that $\delta x$ is small but non-zero.
Again, there are two possibilities:
=== Case 2a ===
If $\delta y = 0$, then $\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = 0$.
Hence the result.
=== Case 2b ===
If $\delta y \ne 0$, then:
:$\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \dfrac {\delta y} {\delta x}$
As $\delta y \to 0$:
:$(1): \quad \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \to \map {f'} y$
:$(2): \quad \dfrac {\delta y} {\delta x} \to 0$
Thus:
:$\ds \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x} \to 0 = \map {f'} y \map {g'} x$
Again, hence the result.
All cases have been covered, so by Proof by Cases, the result is complete.
== Also presented as ==
Category:Derivative of Composite Function