Proof Of The Derivative Of cosh(x)

We need to find the derivative of \(\cosh(x)\):

\[\frac{d}{dx}\cosh(x) = \frac{d}{dx}\frac{e^x + e^{-x}}{2}\]

Since 2 is a constant:

\[\frac{d}{dx}\cosh(x) = \frac{1}{2} \left( \frac{d}{dx} e^x + e^{-x} \right) \]

We can derivative them individually:

\[\begin{align} \frac{d}{dx} \cosh(x) = \frac{1}{2} \left( \frac{d}{dx} e^x + \frac{d}{dx}e^{-x} \right)\\ \frac{d}{dx} \cosh(x) = \frac{1}{2} ( e^x - e^{-x} ) \end{align}\]

This is the definition of \(\sinh(x)\):

\[ \frac{d}{dx} \cosh(x) = \sinh(x)\]

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