Hello wonderful mathematics people, this is Anna Cox from Kellogg community college hyperbolic functions are formed by taking a combination of the two exponential functions e to the x and e to the negative x recall that an even function is f(-x) equaling f(x) or it's symmetric about the y-axis. An odd function f(- x) equals the negative of f(x) is symmetric about the origin so every function that is defined on an interval centered at the origin can be rewritten in a unique way as the sum of an even function and an odd function so we can think of f(x) equaling f(x) plus f(-x) divided by two which was even plus f(x) minus f(-x) divided by two which would be odd so e to the x, we can think of as e to the x plus e to the negative x divided by two plus e to the x minus e to the negative x divided by two. And these are actually going to be definitions the hyperbolic cosine is defined to be, e to the x plus e to the negative x all over two where is the hyperbolic sine is defined to be e to the x minus e to the negative x all over two so we're gonna have some identities the hyperbolic cosine squared x minus the hyperbolic sine squared x is gonna equal one. The proof of this if we have e to the x plus e to the negative x all divided by two squared minus e to the x minus e to the negative x divided by two squared that should equal one so if we distribute we'd get e to the 2x plus two e to the x e to the negative x plus e to the -2x all over four minus, if we distribute the next one and then we realize they're both over the common denominator so we're gonna distribute this negative through and combine our like terms we also know that e to the x times e to the negative x when we have the same bases we're gonna add the exponent so we're gonna get e to the zero and anything to the zero is one so the e to the 2x and the negative e to the 2x are gonna cancel. Two plus two is gonna give us four and e to the -2x and minus e to the -2x are gonna cancel, so four over four which equals one. We just proved that identity other identities that are true that you could prove in similar fashion are hyperbolic sine of 2x is two hyperbolic sine x hyperbolic cosine x hyperbolic cosine of 2x is hyperbolic cosine squared x plus hyperbolic sine squared x hyperbolic cosine squared x equals hyperbolic cosine of 2x plus one all over two hyperbolic sine squared of x equals hyperbolic cosine of 2x minus one over two hyperbolic tangent squared x equals one minus hyperbolic secant squared x hyperbolic cotangent squared x equals one plus hyperbolic cosecant squared x we also know that hyperbolic tangent of x is just gonna be hyperbolic sine over hyperbolic cosine or e to the x minus e to the negative x over e to the x plus e to the negative x the inverse functions occur just like in traditional trig, so the hyperbolic cotangent is just gonna be the reciprocal function of the hyperbolic tangent hyperbolic secant is going to be the reciprocal of hyperbolic cosine hyperbolic cosecant is gonna be the reciprocal of hyperbolic sine so the next thing we're gonna do is we're gonna look at the derivatives of these hyperbolic functions if we have a derivative of hyperbolic sine, we could think of that as e to the x minus e to the negative x over two, because by definition that is what hyperbolic sign is so we're gonna split that up and bring out the one half cause it's a constant so it can come in and out of the derivative at will. The derivative of e to the x is just gonna be e to the x. And the derivative of e to the negative x is negative e to the negative x and the negative of the negative is gonna make that turn into a positive now we have e to the x plus e to the negative x over two and that's really just hyperbolic cosine by definition the derivative of hyperbolic cosine we're gonna find the similar fashion we're gonna end up with one half e to the x minus e to the negative x, so that's gonna be hyperbolic sine when we do hyperbolic tangent think of that as sine divided, hyperbolic sine divided by hyperbolic cosine so the derivative of hyperbolic sine times the cosine, hyperbolic cosine minus the derivative of hyperbolic cosine times the sine all over hyperbolic cosine squared. That was a hyperbolic sine. Remember we're using the quotient rule here so the derivative of hyperbolic sine we just found a minute ago was hyperbolic cosine derivative of hyperbolic cosine was hyperbolic sine we developed a formula of hyperbolic cosine squared minus hyperbolic sine squared is really just one one over hyperbolic cosine squared is hyperbolic secant squared we can do, we have the chain rule for these also so if we're doing hyperbolic sine of some function u it's really just gonna be hyperbolic cosine u du dx. This is true for all six of them if we went ahead and did hyperbolic cotangent, hyperbolic secant, and the hyperbolic cosecant we would find that these are very similar to what we've seen in our regular trig functions with things like negatives sometimes um we need restrict the domain make them all one the one so we have identities of the inverse of hyperbolic secant x is equal to the inverse of hyperbolic cosine of one over x hyperbolic cosecant inverse of x, the inverse function of hyperbolic cosecant of x equals the inverse function of hyperbolic sine of one over x etcetera. We can show that y equal the inverse of hyperbolic cosine x so if I take hyperbolic cosine of each side I'd get hyperbolic cosine of y equaling x now if we take the derivative d dx, we'd get hyperbolic sine y dy dx equaling one. If we divide by hyperbolic sine y we'd get one over hyperbolic sine y. Now we have an identity that says hyperbolic cosine squared y minus hyperbolic sine squared y equal one so we could solve this for hyperbolic sine y. If we have hyperbolic sine y, we'd get that equaling the square root of hyperbolic cosine squared y minus one well if you look at the original, we had y equaling hyperbolic cosine y equaling the inverse function of hyperbolic cosine of x or x is just hyperbolic cosine of y. So right here is hyperbolic cosine of y that's squared so instead of hyperbolic cosine y we're gonna stick in x squared so one over hyperbolic sine y, hyperbolic sine y is just really square root of x squared minus one so if we have y equaling the inverse of hyperbolic cosine x dy dx is just one divided by square root of x squared minus one we're gonna do the same thing for hyperbolic, the inverse function of hyperbolic sine x so we get hyperbolic sine of y equaling x. Taking the derivative of each side hyperbolic cosine y of dy dx equaling one dy dx would be one over hyperbolic cosine of y we have this identity that says hyperbolic cosine squared y minus hyperbolic sine squared y equal. If we solve for hyperbolic cosine of y, because that's what's down here in the denominator we'd see that that's just square root a one plus hyperbolic sine squared y while hyperbolic sine y was x, so that's just one divided by square root of one plus x squared if we look at an example, say we're given hyperbolic cosine of x equaling thirteen fifths where x is greater than zero so we know that the hyperbolic secant functions just gonna be the reciprocal so five thirteenths we know that hyperbolic cosine squared x minus hyperbolic sine squared x is one so if we put and thirteen fifths squared minus one to get our hyperbolic sine squared x. We can solve and get that as twelve fifths. So the hyperbolic cosecant is just the inverse or five twelves the reciprocal hyperbolic tangent as sine, hyperbolic sine divided by hyperbolic cosine or in this case it's gonna be twelve thirteenths hyperbolic cotangent then is thirteen twelfths say we have hyperbolic sine of 2ln(x) so if we wanted to just rewrite this, the first thing we could do is take the two up into the exponent and get hyperbolic sine of ln(x) squared so now using the definition of hyperbolic sine, we'd have e to the ln(x) squared minus e to the negative of ln(x) squared. Now remember again that that negative could go up into the exponent so we'd have ln(x) to the negative two cause two times negative one is just negative two so the e and the ln are gonna cancel and we'd get x squared minus here the e and the ln will cancel again x to the negative two is really just one over x squared divided by two. We're not going to leave a complex fraction so we're gonna multiply x squared over x squared to get x to the fourth minus one divided by two x squared what if we want to find a derivative? y equal t squared hyperbolic tangent of one over t. We're gonna use the product rule and we're also gonna have to use the chain rule. So the derivative of t squared is 2t hyperbolic tangent of one over t. Now the derivative of hyperbolic tangent is hyperbolic secant squared one over t, but then we have to take the derivative of that one over t and it's all gonna get multiplied by the first term of t squared so the derivative of one over t is negative one over t squared the t squared here and the t squared here are gonna cancel, so our finally answer for our first derivative is gonna be y prime equal two t, hyperbolic tangent of one over t minus hyperbolic secant squared of one over t thank you and have a wonderful day