# 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 $x \to \infty$. \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}

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