微分の公式 2 - CANADA'S WINDVIEW

カテゴリー[ 昆虫| 田園| 街| 数学・幾何学| 寺院| 城| 祭り| 鉄道| 海| 風力発電]

イメージ 1

三角関数の微分の公式の導出

三角関数の微分の公式の導出です。

$\displaystyle \lim_{h \to 0} \frac {\sin h}{h} = 1$

$\displaystyle \lim_{h \to 0} \frac {\cos h - 1}{h} = \lim_{h \to 0} \frac {( \cos h - 1 ) ( \cos h + 1 ) h}{h^2 ( \cos h + 1 ) } = \lim_{h \to 0} \frac { \left( \cos^2 h - 1 \right) h}{h^2 ( \cos h + 1 ) } = \lim_{h \to 0} \left( - {\frac {\sin^2 h}{h^2}} \cdot {\frac {1}{ \cos h + 1 }} \cdot h \right) \\ \displaystyle = - 1^2 \cdot {\frac {1}{ 1 + 1 }} \cdot 0 = 0$

上記の結果と加法定理より、

$\displaystyle ( \sin x )' = \lim_{h \to 0} \frac { \sin ( x + h ) - \sin x }{h} = \lim_{h \to 0} \frac { \sin x \cos h + \cos x \sin h - \sin x }{h} \\ \displaystyle = \lim_{h \to 0} \frac { \sin x ( \cos h - 1 ) + \cos x \sin h }{h} = \lim_{h \to 0} \left\{ \sin x \left( \frac {\cos h - 1 }{h} \right) + \cos x \left( \frac {\sin h}{h} \right) \right\} \\ \displaystyle = {\sin x} \cdot 0 + {\cos x} \cdot 1 = \cos x$

$\displaystyle ( \cos x )' = \lim_{h \to 0} \frac { \cos ( x + h ) - \cos x }{h} = \lim_{h \to 0} \frac { \cos x \cos h - \sin x \sin h - \cos x }{h} \\ \displaystyle = \lim_{h \to 0} \frac { \cos x ( \cos h - 1 ) - \sin x \sin h }{h} = \lim_{h \to 0} \left\{ \cos x \left( \frac {\cos h - 1 }{h} \right) - \sin x \left( \frac {\sin h}{h} \right) \right\} \\ \displaystyle = {\cos x} \cdot 0 - {\sin x} \cdot 1 = - \sin x$

上記の結果と商の微分法より、

$\displaystyle ( \tan x )' = \left( \frac {\sin x}{\cos x} \right)' = \frac { ( \sin x )' \cos x - \sin x ( \cos x )' }{\cos^2 x} = \frac { \cos^2 x + \sin^2 x }{\cos^2 x} = \frac { 1 }{\cos^2 x}$

$\displaystyle \left( \frac{ 1 }{ \tan x } \right) ' = \left( \frac{ \cos x }{ \sin x } \right) ' = \frac{ ( \cos x ) ' \sin x - \cos x ( \sin x ) ' }{ \sin^2 x } = \frac{ - \sin^2 x - \cos^2 x }{ \sin^2 x } = - \frac{ 1 }{ \sin^2 x }$

LibreOffice 数式(Math)のソース：

alignl { alignc lim from{ h toward 0 } { { sin h } over { h } } } = 1

alignl { alignc lim from{ h toward 0 } { { cos h - 1 } over { h } } }
= { alignc lim from{ h toward 0 } { { ( cos h - 1 ) ( cos h + 1 ) h } over { h ^2 ( cos h + 1 ) } } }
= { alignc lim from{ h toward 0 } { { ( cos ^2 h - 1 ) h } over { h ^2 ( cos h + 1 ) } } }
= { alignc lim from{ h toward 0 } { left ( - {{ sin ^2 h } over { h ^2 }} cdot {{ 1 } over { cos h + 1 }} cdot h right ) } }
newline
alignl phantom { y } = { alignc { - 1 ^2 } cdot { { 1 } over { 1 + 1 } cdot 0 } } = 0

alignl ( sin x ) ' = { alignc lim from{ h toward 0 } { { sin ( x + h ) - sin x } over { h } } }
= { alignc lim from{ h toward 0 } { { sin x cos h + cos x sin h - sin x } over { h } } }
newline
alignl phantom { y } = { alignc lim from{ h toward 0 } { { sin x ( cos h - 1 ) + cos x sin h } over { h } } } = { alignc lim from{ h toward 0 } { left lbrace sin x left ( { cos h - 1 } over { h } right ) + cos x left ( { sin h } over { h } right ) right rbrace } } = cos x

alignl ( cos x ) ' = { alignc lim from{ h toward 0 } { { cos ( x + h ) - cos x } over { h } } }
= { alignc lim from{ h toward 0 } { { cos x cos h - sin x sin h - cos x } over { h } } }
newline
alignl phantom { y } = { alignc lim from{ h toward 0 } { { cos x ( cos h - 1 ) - sin x sin h } over { h } } } = { alignc lim from{ h toward 0 } { left lbrace cos x left ( { cos h - 1 } over { h } right ) - sin x left ( { sin h } over { h } right ) right rbrace } } = - sin x

alignl ( tan x ) ' = left ( { sin x } over { cos x } right ) ' = { alignc { ( sin x )' cos x - sin x ( cos x )' } over { cos ^2 x } } = { alignc { cos ^2 x + sin ^2 x } over { cos ^2 x } } = { alignc { 1 } over { cos ^2 x } }