MA101.14 The hyperbolic functions

We define:

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

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

Note that:

\sinh{-x}=\frac{1}{2}(e^{-x}-e^x)=-\sinh{x} – so an odd function.

\cosh{-x}=\frac{1}{2}(e^{-x}+e^x)=\cosh{x} – so an even function.

Also: \cosh{x}-\sinh{x}=e^{-x}\rightarrow 0 as x \to \infty.

We also define:

\begin{aligned} \tanh{x}&=\frac{\sinh{x}}{\cosh{x}} \\ \coth{x}&=\frac{\cosh{x}}{\sinh{x}}, x\neq 0 \\ \mbox{sech }x &=\frac{1}{\cosh{x}} \\ \mbox{cosech }x &=\frac{1}{\sinh{x}}, x\neq 0 \end{aligned}

Note:

\begin{aligned} \tanh{x} &= \frac{e^x-e^{-x}}{e^x+e^{-x}} \\&= \frac{1-e^{-2x}}{1+e^{-2x}} \rightarrow 1 \;\mbox{from below as} x\to\infty \\ &= \frac{e^{2x}-1}{e^{2x}+1} \rightarrow -1 \;\mbox{from above as} x\to -\infty \end{aligned}

Derivative of sinh and cosh

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

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

Exercise
Find derivatives of tanh, coth, sech,cosech.

Inverse functions

sinh is an injection and so we have an inverse function denoted by \sinh^{-1}{x}.

Domain \sinh^{-1}{x} = range of \sinh{x} = \mathbb{R}
Range \sinh^{-1}{x} = domain of \sinh{x} = \mathbb{R}

\sinh^{-1}{x} means the real number whose \sinh{x} is x.

tanh is an injection and so we have an inverse function denoted by \tanh^{-1}{x}.

Domain \tanh^{-1}{x} = range of \tanh{x} = (-1,1)
Range \tanh^{-1}{x} = domain of \tanh{x} = \mathbb{R}

\tanh^{-1}{x} means the real number whose \tanh{x} is x.

Note that \cosh{x} is not an injection, so we need a cut down domain to non-negative x’s (x \geq 0) to make it one.

\cosh^{-1}{x} means the non-negative real number whose \cosh{x} is x.

Derivatives of the inverse hyperbolic functions

\begin{aligned} y &= \cosh^{-1}{x} \\ x &= \cosh{y} \\ \frac{dx}{dy}&=\sinh{y} \\ \frac{dy}{dx} &= \frac{1}{\sinh{y}} \\ &= \frac{1}{\sqrt{x^1-1}} \end{aligned}

\mbox{Note: }\forall x: \cosh{^2}{x}-\sinh{^2}{x}=1.

\begin{aligned} y &= \sinh^{-1}{x} \\ x & = \sinh{y} \\ \frac{dx}{dy} &= \cosh{y} \\ \frac{dy}{dx} &= \frac{1}{\cosh{y}}  \\&= \frac{1}{\sqrt{1+x^2}} \end{aligned}