三角関数と双曲線関数の微分
三角関数の微分
(1)
\[ \frac{d}{dx}\sin x=\cos x \](2)
\[ \frac{d}{dx}\cos x=-\sin x \](3)
\[ \frac{d}{dx}\tan x=\cos^{-2}x \](4)
\[ \frac{d}{dx}\sin^{-1}x=-\sin^{-1}x\tan^{-1}x \](5)
\[ \frac{d}{dx}\cos^{-1}x=\cos^{-1}x\tan x \](6)
\[ \frac{d}{dx}\tan^{-1}x=-\sin^{-2}x \](1)
\begin{align*} \frac{d}{dx}\sin x & =\frac{d}{dx}\frac{e^{ix}-e^{-ix}}{2i}\\ & =\frac{ie^{ix}+ie^{-ix}}{2i}\\ & =\cos x \end{align*}(2)
\begin{align*} \frac{d}{dx}\cos x & =\frac{d}{dx}\frac{e^{ix}+e^{-ix}}{2}\\ & =\frac{ie^{ix}-ie^{-ix}}{2}\\ & =-\frac{e^{ix}-e^{-ix}}{2i}\\ & =-\sin x \end{align*}(3)
\begin{align*} \frac{d}{dx}\tan x & =\frac{d}{dx}\frac{\sin x}{\cos x}\\ & =\frac{(\sin x)'\cos x-\sin x(\cos x)'}{\cos^{2}x}\\ & =\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}\\ & =\cos^{-2}x \end{align*}(4)
\begin{align*} \frac{d}{dx}\sin^{-1}x & =\frac{d\sin x}{dx}\frac{d\sin^{-1}x}{d\sin x}\\ & =\cos x(-\sin^{-2}x)\\ & =-\sin^{-1}x\tan^{-1}x \end{align*}(5)
\begin{align*} \frac{d}{dx}\cos^{-1}x & =\frac{d\cos x}{dx}\frac{d\cos^{-1}x}{d\cos x}\\ & =-\sin x(-\cos^{-2}x)\\ & =\cos^{-1}x\tan x \end{align*}(6)
\begin{align*} \frac{d}{dx}\tan^{-1}x & =\frac{d\tan x}{dx}\frac{d\tan^{-1}x}{d\tan x}\\ & =\cos^{-2}x(-\tan^{-2}x)\\ & =-\sin^{-2}x \end{align*}双曲線関数の微分
(1)
\[ \frac{d}{dx}\sinh x=\cosh x \](2)
\[ \frac{d}{dx}\cosh x=\sinh x \](3)
\[ \frac{d}{dx}\tanh x=\cosh^{-2}x \](4)
\[ \frac{d}{dx}\sinh^{-1}x=-\sinh^{-1}x\tanh^{-1}x \](5)
\[ \frac{d}{dx}\cosh^{-1}x=-\cosh^{-1}x\tanh x \](6)
\[ \frac{d}{dx}\tanh^{-1}x=-\sinh^{-2}x \](1)
\begin{align*} \frac{d}{dx}\sinh x & =-i\frac{d}{dx}\sin(ix)\\ & =\cos(ix)\\ & =\cosh x \end{align*}(2)
\begin{align*} \frac{d}{dx}\cosh x & =\frac{d}{dx}\cos(ix)\\ & =-i\sin(ix)\\ & =\sinh x \end{align*}(3)
\begin{align*} \frac{d}{dx}\tanh x & =-i\frac{d}{dx}\tan(ix)\\ & =\cos^{-2}(ix)\\ & =\cosh^{-2}x \end{align*}(4)
\begin{align*} \frac{d}{dx}\sinh^{-1}x & =i\frac{d}{dx}\sin^{-1}(ix)\\ & =\sin^{-1}(ix)\tan^{-1}(ix)\\ & =-\sinh^{-1}x\tanh^{-1}x \end{align*}(5)
\begin{align*} \frac{d}{dx}\cosh^{-1}x & =\frac{d}{dx}\cos^{-1}(ix)\\ & =i\cos^{-1}(ix)\tan(ix)\\ & =-\cosh^{-1}x\tanh x \end{align*}(6)
\begin{align*} \frac{d}{dx}\tanh^{-1}x & =i\frac{d}{dx}\tan^{-1}(ix)\\ & =\sin^{-2}(ix)\\ & =-\sinh^{-2}x \end{align*}ページ情報
タイトル | 三角関数と双曲線関数の微分 |
URL | https://www.nomuramath.com/xw1wbjdm/ |
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三角関数の部分分数展開
\[
\pi\tan\pi x =-\sum_{k=-\infty}^{\infty}\frac{1}{x+\frac{1}{2}+k}
\]
三角関数と双曲線関数の積分
\[
\int f(\cos x,\sin x)dx=\int f\left(\frac{1-t^{2}}{1+t^{2}},\frac{2t}{1+t^{2}}\right)\frac{2}{1+t^{2}}dt\cmt{t=\tan\frac{x}{2}}
\]
三角関数と双曲線関数の冪乗積分漸化式
\[
\int\sin^{n}xdx=-\frac{1}{n}\cos x\sin^{n-1}x+\frac{n-1}{n}\int\sin^{n-2}xdx\qquad(n\ne0)
\]
三角関数と双曲線関数の対数の積分
\[
\int\Log\sin^{\alpha}zdz=z\Log\sin^{\alpha}x+\frac{i\alpha}{2}z^{2}+\alpha z\Li_{1}\left(e^{2iz}\right)+\frac{i\alpha}{2}\Li_{2}\left(e^{2iz}\right)+\C{}
\]