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Prove that the derivative of cosh x is sinh x

Prove the following formula for the derivative of hyperbolic cosine,

    \[ D (\cosh x) = \sinh x. \]


Proof. We can compute from the definition of hyperbolic cosine in terms of exponentials,

    \begin{align*}  D(\cosh x) &= D \left( \frac{e^x + e^{-x}}{2} \right) \\  &= \frac{1}{2} (e^x - e^{-x}) \\  &= \sinh x. \qquad \blacksquare \end{align*}

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