$$ \newcommand\bm[1]{\boldsymbol{#1}} \renewcommand\limsup{\varlimsup} \renewcommand\liminf{\varliminf} $$

次の等式を証明せよ. \begin{equation*} \begin{split} &\frac{d}{dx}\sinh{x} = \cosh{x}, \\ &\frac{d}{dx}\cosh{x} = \sinh{x}, \\ &\frac{d}{dx}\tanh{x} = \frac{1}{\cosh^2{x}}. \end{split} \end{equation*}

解答例 1

\begin{align*} \frac{d}{dx}\sinh{x} &= \frac{d}{dx}\left(\frac{e^{x}-e^{-x}}{2}\right) = \frac{e^{x}+e^{-x}}{2} = \cosh{x}, \\ \frac{d}{dx}\cosh{x} &= \frac{d}{dx}\left(\frac{e^{x}+e^{-x}}{2}\right) = \frac{e^{x}-e^{-x}}{2} = \sinh{x}. \end{align*} さらに, \begin{align*} \frac{d}{dx}\tanh{x} &= \frac{d}{dx}\left(\frac{\sinh{x}}{\cosh{x}}\right) \\ &= \frac{(\sinh{x})'\cosh{x}-\sinh{x}(\cosh{x})'}{\cosh^2{x}} \\ &= \frac{\cosh^2{x}-\sinh^2{x}}{\cosh^2{x}} = \frac{1}{\cosh^2{x}}. \end{align*}

最終更新日:2011年11月02日

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