逆三角関数と逆双曲線関数の微分
逆三角関数の微分
(1)
\[ \frac{d}{dx}\sin^{\bullet}x=\frac{1}{\sqrt{1-x^{2}}} \](2)
\[ \frac{d}{dx}\cos^{\bullet}x=\frac{-1}{\sqrt{1-x^{2}}} \](3)
\[ \frac{d}{dx}\tan^{\bullet}x=\frac{1}{1+x^{2}} \](4)
\[ \frac{d}{dx}\sin^{-1,\bullet}x=\frac{-1}{x^{2}\sqrt{1-x^{-2}}} \](5)
\[ \frac{d}{dx}\cos^{-1,\bullet}x=\frac{1}{x^{2}\sqrt{1-x^{-2}}} \](6)
\[ \frac{d}{dx}\tan^{-1,\bullet}x=\frac{-1}{1+x^{2}} \](1)
\begin{align*} \frac{d}{dx}\sin^{\bullet}x & =\frac{d\sin^{\bullet}x}{d\sin\sin^{\bullet}x}\\ & =\cos^{-1}\sin^{\bullet}x\\ & =\frac{1}{\sqrt{1-\sin^{2}\sin^{\bullet}x}}\\ & =\frac{1}{\sqrt{1-x^{2}}} \end{align*}(2)
\begin{align*} \frac{d}{dx}\cos^{\bullet}x & =\frac{d\cos^{\bullet}x}{d\cos\cos^{\bullet}x}\\ & =-\sin^{-1}\cos^{\bullet}x\\ & =\frac{-1}{\sqrt{1-\cos^{2}\cos^{\bullet}x}}\\ & =\frac{-1}{\sqrt{1-x^{2}}} \end{align*}(3)
\begin{align*} \frac{d}{dx}\tan^{\bullet}x & =\frac{d\tan^{\bullet}x}{d\tan\tan^{\bullet}x}\\ & =\cos^{2}\tan^{\bullet}x\\ & =\frac{1}{\sqrt{1+\tan^{2}\tan^{\bullet}x}}\\ & =\frac{1}{1+x^{2}} \end{align*}(4)
\begin{align*} \frac{d}{dx}\sin^{-1,\bullet}x & =\frac{d\sin^{-1,\bullet}x}{d\sin^{-1}\sin^{-1,\bullet}x}\\ & =-\sin^{2}\sin^{-1,\bullet}x\cos^{-1}\sin^{-1,\bullet}x\\ & =\frac{-1}{x^{2}\sqrt{1-\sin^{2}\sin^{-1,\bullet}x}}\\ & =\frac{-1}{x^{2}\sqrt{1-x^{-2}}} \end{align*}(5)
\begin{align*} \frac{d}{dx}\cos^{-1,\bullet}x & =\frac{d\cos^{-1,\bullet}x}{d\cos^{-1}\cos^{-1,\bullet}x}\\ & =\cos^{2}\cos^{-1,\bullet}x\sin^{-1}\cos^{-1,\bullet}x\\ & =\frac{1}{x^{-2}\sqrt{1-\cos^{2}\cos^{-1,\bullet}x}}\\ & =\frac{1}{x^{2}\sqrt{1-x^{-2}}} \end{align*}(6)
\begin{align*} \frac{d}{dx}\tan^{-1,\bullet}x & =\frac{d\tan^{-1,\bullet}x}{d\tan^{-1}\tan^{-1,\bullet}x}\\ & =-\sin^{2}\tan^{-1,\bullet}x\\ & =\frac{-1}{\sqrt{1+\tan^{-2}\tan^{-1,\bullet}x}}\\ & =\frac{-1}{1+x^{2}} \end{align*}逆双曲線関数の微分
(1)
\[ \frac{d}{dx}\sinh^{\bullet}x=\frac{1}{\sqrt{1+x^{2}}} \](2)
\[ \frac{d}{dx}\cosh^{\bullet}x=\frac{1}{\sqrt{x^{2}-1}} \](3)
\[ \frac{d}{dx}\tanh^{\bullet}x=\frac{1}{1-x^{2}} \](4)
\[ \frac{d}{dx}\sinh^{-1,\bullet}x=\frac{-1}{x^{2}\sqrt{1+x^{-2}}} \](5)
\[ \frac{d}{dx}\cosh^{-1,\bullet}x=\frac{-1}{x^{2}\sqrt{x^{-2}-1}} \](6)
\[ \frac{d}{dx}\tanh^{-1,\bullet}x=\frac{1}{1-x^{2}} \](1)
\begin{align*} \frac{d}{dx}\sinh^{\bullet}x & =\frac{d\sinh^{\bullet}x}{d\sinh\sinh^{\bullet}x}\\ & =\cosh^{-1}\sinh^{\bullet}x\\ & =\frac{1}{\sqrt{1+\sinh^{2}\sinh^{\bullet}x}}\\ & =\frac{1}{\sqrt{1+x^{2}}} \end{align*}(2)
\begin{align*} \frac{d}{dx}\cosh^{\bullet}x & =\frac{d\cosh^{\bullet}x}{d\cosh\cosh^{\bullet}x}\\ & =\sinh^{-1}\cosh^{\bullet}x\\ & =\frac{1}{\sqrt{\cosh^{2}\cosh^{\bullet}x-1}}\\ & =\frac{1}{\sqrt{x^{2}-1}} \end{align*}(3)
\begin{align*} \frac{d}{dx}\tanh^{\bullet}x & =\frac{d\tanh^{\bullet}x}{d\tanh\tanh^{\bullet}x}\\ & =\cosh^{2}\tanh^{\bullet}x\\ & =\frac{1}{1-\tanh^{2}\tanh^{\bullet}x}\\ & =\frac{1}{1-x^{2}} \end{align*}(4)
\begin{align*} \frac{d}{dx}\sinh^{-1,\bullet}x & =\frac{d\sinh^{-1,\bullet}x}{d\sinh^{-1}\sinh^{-1,\bullet}x}\\ & =-\sinh^{2}\sinh^{-1,\bullet}x\cosh^{-1}\sinh^{-1,\bullet}x\\ & =\frac{-1}{x^{2}\sqrt{1+\sinh^{2}\sinh^{-1,\bullet}x}}\\ & =\frac{-1}{x^{2}\sqrt{1+x^{-2}}} \end{align*}(5)
\begin{align*} \frac{d}{dx}\cosh^{-1,\bullet}x & =\frac{d\cosh^{-1,\bullet}x}{d\cosh^{-1}\cosh^{-1,\bullet}x}\\ & =-\cosh^{2}\cosh^{-1,\bullet}x\sinh^{-1}\cosh^{-1,\bullet}x\\ & =\frac{-1}{x^{2}\sqrt{\cosh^{2}\cosh^{-1,\bullet}x-1}}\\ & =\frac{-1}{x^{2}\sqrt{x^{-2}-1}} \end{align*}(6)
\begin{align*} \frac{d}{dx}\tanh^{-1,\bullet}x & =\frac{d\tanh^{-1,\bullet}x}{d\tanh^{-1}\tanh^{-1,\bullet}x}\\ & =-\tanh^{2}\tanh^{-1,\bullet}x\cosh^{2}\tanh^{-1,\bullet}x\\ & =\frac{-1}{x^{2}\left(1-\tanh^{2}\tanh^{-1,\bullet}x\right)}\\ & =\frac{1}{1-x^{2}} \end{align*}ページ情報
| タイトル | 逆三角関数と逆双曲線関数の微分 |
| URL | https://www.nomuramath.com/aakd9f6o/ |
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三角関数・双曲線関数の一次結合の逆数の積分
\[
\int\frac{1}{\alpha\sin z+\beta\cos z+\gamma}dz=-\frac{2}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}\tanh^{\bullet}\frac{\left(\gamma-\beta\right)\tan\frac{z}{2}+\alpha}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}+C
\]
三角関数・双曲線関数の微分
\[
\left(\sin x\right)'=\cos x
\]
逆三角関数と逆双曲線関数の関係
\[
\Sin^{\bullet}\left(iz\right)=i\Sinh^{\bullet}z
\]
1と3角関数・双曲線関数
\[
1+\sin z=\left(\cos\frac{z}{2}+\sin\frac{z}{2}\right)^{2}
\]