Why hyperbolics function are called 'hyperbolic'? I am studying now calculus and I am learning about hyperbolic function.
But I can't understand why hyperbolic functions are called 'hyperbolics'
Is there any relation between hyperbolic function and hyperbola?
And why there are natural constant e in function?
I saw some information that notices the relation between them in Wikipedia, so I try to find the relation as integrating the hyperbola, but I can't solve that.
Please tell me why they are called 'hyperbolic' and derivate the hyperbolic function.
(If the hyperbolic function is just defined so, please tell me why they are defined so.)
 A: The "trigonomic" functions, sine, cosine, etc.,  in Calculus are often called "circular functions" because, in order to avoid reference to units, like "degrees", and to be able to extend beyond 0 to 90 degrees we redefine them:  Given a circle with center at (0, 0) on an xy-coordinate system, radius 1,  start at (1, 0) and measure around the circle a distance $\theta$.  We define "$\cos(\theta)$" and "$\sin(\theta)$" as the x and y coordinates of the end point.
That can be done because $sin^2(\theta)+ cos^2(\theta)= 1$.  It is just the $x^2+ y^2= 1$ of the unit circle.
For the "hyperbolic functions", start with an xy-coordinate system and drew the hyperbolic graph, $x^2- y^2= 1$.  Starting from (1, 0) on that graph measure a distance "t" along the graph.  "cosh(t)" and "sinh(t)" are defined as the (x, y) coordinates of the endpoint and, of course, $cosh^2(t)- sinh^2(t)= 1$.
A: Of the 3 main hyperbolic functions $\sinh(x)$, $\cosh(x)$, and $\tanh(x)$ (which I'm going to talk about because they're the most common ones), only $\cosh(x)$ is a hyperbola. I don't really know why e is in them. $$\frac{d}{dx}\sinh(x)=\frac12\frac{d}{dx}e^x-e^{-x}=\frac12e^x-(-1)e^{-x}=\cosh(x)$$$$\frac{d}{dx}\cosh(x)=\frac12\frac{d}{dx}e^x+e^{-x}=\frac12e^x+(-1)e^{-x}=\sinh(x)$$$$\frac{d}{dx}\tanh(x)=\frac{d}{dx}\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{(e^x+e^{-x})(e^x+e^{-x})-(e^x-e^{-x})(e^x-e^{-x})}{(e^x-e^{-x})^2}=\frac{1}{\cosh^2(x)}=sech^2(x)$$Note: As $(\frac{e^x\pm e^{-x}}{2})^2=\frac{e^{2x}\pm 2*e^x*e^{-x}+e^{-2x}}{2^2}=\frac{e^{2x}+e^{-2x}\pm 2*e^0}{4}=\frac{e^{2x}+e^{-2x}}{4}\pm\frac12$. Because of this, $\cosh^2(x)-\sinh^2(x)=0$.
