📂Functions

Derivatives of Hyperbolic Functions

Formula¹

The derivative of hyperbolic functions is as follows:

$$\begin{align*} \dfrac{d}{dx} (\sinh x) &= \cosh x \qquad & \dfrac{d}{dx} (\csch x) &= - \csch x \coth x \\ \dfrac{d}{dx} (\cosh x) &= \sinh x \qquad & \dfrac{d}{dx} (\sech x) &= - \sech x \tanh x \\ \dfrac{d}{dx} (\tanh x) &= \sech^{2} x \qquad & \dfrac{d}{dx} (\coth x) &= - \csch^{2} x \end{align*}$$

Proof

Since hyperbolic functions are linear combinations of exponential functions, their derivatives can be easily found.

$(\sinh x)^{\prime}$, $(\cosh x)^{\prime}$

$$\begin{align*} \dfrac{d}{dx} \sinh x &= \dfrac{d}{dx} \left( \dfrac{e^{x} - e^{-x}}{2} \right) \\ &= \dfrac{e^{x} + e^{-x}}{2} \\ &= \cosh x \end{align*}$$

$$\begin{align*} \dfrac{d}{dx} \cosh x &= \dfrac{d}{dx} \left( \dfrac{e^{x} + e^{-x}}{2} \right) \\ &= \dfrac{e^{x} - e^{-x}}{2} \\ &= \sinh x \end{align*}$$

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$(\tanh x)^{\prime}$, $(\coth x)^{\prime}$

Using the quotient rule and hyperbolic identities $\cosh^{2} - \sinh^{2} =1$, it can be derived.

$$\begin{align*} \dfrac{d}{dx} \tanh x &= \dfrac{d}{dx} \left( \dfrac{\sinh x}{\cosh x} \right) \\[1em] &= \dfrac{(\sinh x)^{\prime} \cosh x - \sinh x (\cosh x)^{\prime}}{\cosh^{2} x} \\[1em] &= \dfrac{\cosh x \cosh x - \sinh x \sinh x}{\cosh^{2} x} \\[1em] &= \dfrac{\cosh^{2} x - \sinh^{2} x}{\cosh^{2} x} \\[1em] &= \dfrac{1}{\cosh^{2} x} \\[1em] &= \sech^{2} x \end{align*}$$

$$\begin{align*} \dfrac{d}{dx} \coth x &= \dfrac{d}{dx} \left( \dfrac{\cosh x}{\sinh x} \right) \\[1em] &= \dfrac{(\cosh x)^{\prime} \sinh x - \cosh x (\sinh x)^{\prime}}{\sinh^{2} x} \\[1em] &= \dfrac{\sinh x \sinh x - \cosh x \cosh x}{\sinh^{2} x} \\[1em] &= \dfrac{\sinh^{2} x - \cosh^{2} x}{\sinh^{2} x} \\[1em] &= -\dfrac{1}{\sinh^{2} x} \\[1em] &= -\csch^{2} x \end{align*}$$

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$(\csch x)^{\prime}$, $(\sech x)^{\prime}$

By the chain rule,

$$\begin{align*} \dfrac{d}{dx}\csch x &= \dfrac{d}{dx} \left( \dfrac{1}{\sinh x} \right) \\[1em] &= -\dfrac{1}{\sinh^{2} x} (\sinh x)^{\prime} \\[1em] &= -\dfrac{1}{\sinh^{2} x} \cosh x \\[1em] &= -\csch^{2} x \cosh x \\ &= -\csch x \coth x \end{align*}$$

$$\begin{align*} \dfrac{d}{dx}\sech x &= \dfrac{d}{dx} \left( \dfrac{1}{\cosh x} \right) \\[1em] &= -\dfrac{1}{\cosh^{2} x} (\cosh x)^{\prime} \\[1em] &= -\dfrac{1}{\cosh^{2} x} \sinh x \\[1em] &= -\sech^{2} x \sinh x \\ &= -\sech x \tanh x \end{align*}$$

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