Hyperbolic functions are analogs of the ordinary trigonometric functions, but defined using the hyperbola rather than the circle. They are defined in terms of the exponential function $e^x$.

• Hyperbolic Sine (sinh): $\sinh x = \frac{e^x - e^{-x}}{2}$
• Hyperbolic Cosine (cosh): $\cosh x = \frac{e^x + e^{-x}}{2}$
• Hyperbolic Tangent (tanh): $\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$
• Hyperbolic Cosecant (csch): $\text{csch } x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}$, for $x \neq 0$
• Hyperbolic Secant (sech): $\text{sech } x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$
• Hyperbolic Cotangent (coth): $\coth x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$, for $x \neq 0$

A fundamental identity, analogous to $\sin^2 x + \cos^2 x = 1$, is $\cosh^2 x - \sinh^2 x = 1$.

The derivatives of hyperbolic functions can be found by differentiating their exponential definitions, using the fact that $\frac{d}{dx}(e^x) = e^x$ and $\frac{d}{dx}(e^{-x}) = -e^{-x}$.

Derivative of $\sinh x$

$\frac{d}{dx}(\sinh x) = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{1}{2} \left(\frac{d}{dx}(e^x) - \frac{d}{dx}(e^{-x})\right)$

$= \frac{1}{2} (e^x - (-e^{-x})) = \frac{e^x + e^{-x}}{2} = \cosh x$

So, $\frac{d}{dx}(\sinh x) = \cosh x$.

Derivative of $\cosh x$

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

$= \frac{1}{2} (e^x + (-e^{-x})) = \frac{e^x - e^{-x}}{2} = \sinh x$

So, $\frac{d}{dx}(\cosh x) = \sinh x$.

Derivative of $\tanh x$

Using the quotient rule for $\tanh x = \frac{\sinh x}{\cosh x}$:

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

$= \frac{(\cosh x)(\cosh x) - (\sinh x)(\sinh x)}{\cosh^2 x} = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x}$

Since $\cosh^2 x - \sinh^2 x = 1$,

$= \frac{1}{\cosh^2 x} = \text{sech}^2 x$

So, $\frac{d}{dx}(\tanh x) = \text{sech}^2 x$.

Derivative of $\text{csch } x$

Using the chain rule for $\text{csch } x = (\sinh x)^{-1}$:

$\frac{d}{dx}(\text{csch } x) = -1 (\sinh x)^{-2} \cdot \frac{d}{dx}(\sinh x) = -(\sinh x)^{-2} (\cosh x)$

$= -\frac{\cosh x}{\sinh^2 x} = -\frac{1}{\sinh x} \cdot \frac{\cosh x}{\sinh x} = -\text{csch } x \coth x$

So, $\frac{d}{dx}(\text{csch } x) = -\text{csch } x \coth x$.

Derivative of $\text{sech } x$

Using the chain rule for $\text{sech } x = (\cosh x)^{-1}$:

$\frac{d}{dx}(\text{sech } x) = -1 (\cosh x)^{-2} \cdot \frac{d}{dx}(\cosh x) = -(\cosh x)^{-2} (\sinh x)$

$= -\frac{\sinh x}{\cosh^2 x} = -\frac{1}{\cosh x} \cdot \frac{\sinh x}{\cosh x} = -\text{sech } x \tanh x$

So, $\frac{d}{dx}(\text{sech } x) = -\text{sech } x \tanh x$.

Derivative of $\coth x$

Using the quotient rule for $\coth x = \frac{\cosh x}{\sinh x}$:

$\frac{d}{dx}(\coth x) = \frac{\frac{d}{dx}(\cosh x) \cdot \sinh x - \cosh x \cdot \frac{d}{dx}(\sinh x)}{\sinh^2 x}$

$= \frac{(\sinh x)(\sinh x) - (\cosh x)(\cosh x)}{\sinh^2 x} = \frac{\sinh^2 x - \cosh^2 x}{\sinh^2 x}$

Since $\cosh^2 x - \sinh^2 x = 1 \implies \sinh^2 x - \cosh^2 x = -1$,

$= \frac{-1}{\sinh^2 x} = -\text{csch}^2 x$

So, $\frac{d}{dx}(\coth x) = -\text{csch}^2 x$.

If $u$ is a differentiable function of $x$, the chain rule applies as follows:

• $\frac{d}{dx}(\sinh u) = (\cosh u) \cdot \frac{du}{dx}$
• $\frac{d}{dx}(\cosh u) = (\sinh u) \cdot \frac{du}{dx}$
• $\frac{d}{dx}(\tanh u) = (\text{sech}^2 u) \cdot \frac{du}{dx}$
• $\frac{d}{dx}(\text{csch } u) = (-\text{csch } u \coth u) \cdot \frac{du}{dx}$
• $\frac{d}{dx}(\text{sech } u) = (-\text{sech } u \tanh u) \cdot \frac{du}{dx}$
• $\frac{d}{dx}(\coth u) = (-\text{csch}^2 u) \cdot \frac{du}{dx}$