Proof Of The Derivative Of tanh(x)

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

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

We can use the qoutient rule:

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

After the derivation:

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

Since \(\cosh^2(x) - \sinh^2(x) = 1\):

\[\frac{d}{dx}\tanh(x) = \frac{1}{\cosh^2(x)} = \operatorname{sech}^2(x) \]

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