📂Calculus

Differentiation of Hyperbolic Functions

Theorem¹

$$\left( \sinh x \right)^{\prime} = \cosh x$$

$$\left( \cosh x \right)^{\prime} = \sinh x$$

$$\left( \tanh x \right)^{\prime} = \text{sech}^{2} x$$

Explanation

The differentiation of hyperbolic functions actually doesn’t require much proof or memorization. The proofs simply use definitions, and the structures are almost identical to trigonometric functions, only with a change in signs. Using the method of proving hyperbolic sine, one can easily obtain the derivative of hyperbolic cosine. The derivative of the hyperbolic tangent is derived by applying the quotient rule of differentiation.

Proof

$\sinh$

Since $\sinh x = {{e^x - e^{-x}} \over {2}}$,

$$\left( \sinh x \right)^{\prime} = {{e^x - (-1) e^{-x}} \over {2}} = {{e^x + e^{-x}} \over {2}} = \cosh x$$

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$\tanh$

$$\left( \tanh x \right)^{\prime} = \left( { {\sinh x} \over {\cosh x } } \right)^{\prime}$$

According to the differential formulas obtained earlier, $\left( \sinh x \right)^{\prime} = \cosh x$ and $\left( \cosh x \right)^{\prime} = \sinh x$,

$$\left( { {\sinh x} \over {\cosh x } } \right)^{\prime} = { {\cosh ^{2} x - \sinh ^{2} x} \over {\cosh ^{2} x } } = \text{sech} ^{2} x$$

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