Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. Hyperbolic functions are formed by taking combination of the two exponential functions E to the X&E to the negative X. Recall that an even function is F of negative X equaling F of X, or it's symmetric about the Y axis and odd function F of negative X equals the negative of F of 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 of X equaling F of X + F of negative X / 2, which was even, plus F of X -, F of negative X / 2, which would be odd. So E to the X we could think of as E to the X + e to the next E to the negative X / 2 + e to the X -, e to the negative X / 2. And these are actually going to be definitions. The hyperbolic cosine is defined to be E to the X + e to the negative X all over 2, whereas the hyperbolic sine is defined to be E to the X -, e to the negative X all over 2. So we're going to have some identities. The hyperbolic cosine squared X minus the hyperbolic sine squared X is going to equal 1. The proof of this, if we have E to the X + e to the negative X all divided by 2 ^2 -, e to the X -, e to the negative X / 2 ^2, that should equal 1. So if we distribute, we'd get E to the two X + 2, E to the xe to the negative X + e to the -2 X all over 4 minus if we distribute the next one, and then we realize they're both over the common denominator. So we're going to distribute this negative through and combine our like terms. We also know that E to the X * e to the negative X. When we have the same basis, we're going to add the exponents. So we're going to get E to the zero and anything to the zero is 1. So the E to the 2X and the negative E to the 2X are going to cancel. 2 + 2 is going to give us four and E to the -2 X and minus E to the -2 X are going to cancel. So 4 / 4 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 2 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 2 hyperbolic sine squared of X equals hyperbolic cosine of two X -. 1 / 2 hyperbolic tangent squared X = 1 minus hyperbolic secant squared X hyperbolic cotangent squared X = 1 plus hyperbolic cosecant squared X. We also know that hyperbolic tangent of X is just going to be hyperbolic sine over hyperbolic cosine or E to the X -, e to the negative X / e to the X + e to the negative X. The inverse functions occur just like in traditional trig, so the hyperbolic cotangent is just going to be the reciprocal function of the hyperbolic tangent. Hyperbolic secant is going to be the reciprocal of hyperbolic cosine. Hyperbolic cosecant is going to be the reciprocal of hyperbolic sine. So the next thing we're going to do is we're going to 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 -, e to the negative X / 2, because by definition that is what hyperbolic sine is. So we're going to split that up. We're going to bring out the 1/2 because 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 going to 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 going to make that turn into a positive. Now we have E to the X + e to the negative X / 2, and that's really just hyperbolic cosine. By definition, the derivative of hyperbolic cosine, we're going to find the similar fashion. We're going to end up with 1/2 E to the X -, e to the negative X. So that's going to 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 Cos 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 could do we have the chain rule for these also. So if we're doing hyperbolic sine of some function U, it's really just going to be hyperbolic cosine UDUDX. This is true for all six of them. If we went ahead and did hyperbolic cotangent, hyperbolic secant, and 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 we need to restrict the domain to make them all one to 1. So we have identities of the inverse of hyperbolic secant X is equal to the inverse of hyperbolic cosine of 1 / X, hyperbolic cosecant inverse of X. The inverse function of hyperbolic cosecant of X equals the inverse function of hyperbolic sine of 1 / 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 YDY 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 1. 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 -, 1. 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 going to stick in X ^2. So one over hyperbolic sine Y hyperbolic sine Y is just really sqrt X ^2 - 1. So if we have Y equaling the inverse of hyperbolic cosine XDYDX is just 1 / sqrt X ^2 - 1, We're going to 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 1. 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 1. If we solved for hyperbolic cosine of Y because that's what's down here in the denominator, we'd see that that's just sqrt 1 plus hyperbolic sine squared Y. Well, hyperbolic sine Y was X, so that's just 1 / sqrt 1 + X ^2. If we look at an example, say we're given hyperbolic cosine of X equaling 13 fifths where X is greater than 0. So we know that the hyperbolic secant functions just going to be the reciprocal SO5 thirteenths. We know that hyperbolic cosine squared X minus hyperbolic sine squared X is 1. So if we put in 13 fifth squared -1 to get our hyperbolic sine squared X, we can solve and get that as 12 fifths. So the hyperbolic cosecant is just the inverse or five twelfths. The reciprocal hyperbolic tangent as sine. Hyperbolic sine divided by hyperbolic cosine, or in this case it's going to be 12 thirteenths. Hyperbolic cotangent then is 13 twelfths. Say we have hyperbolic sine of two LNX. 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 LNX squared. So now using the definition of hyperbolic sine, we'd have E to the LNX squared minus E to the negative of LNX squared. Now remember again that that negative could go up into the exponent, so we'd have LNX to the -2 because 2 * -1 is just -2. So the E and the lane are going to cancel and we get X ^2 minus. Here the E and the lane will cancel again. X to the -2 is really just 1 / X ^2 divided by two. We're not going to leave a complex fraction, so we're going to multiply X ^2 / X ^2 to get X to the 4th -1 / 2 X squared. What if we want to find a derivative Y equal t ^2 hyperbolic tangent of 1 / t? We're going to use the product rule, and we're also going to have to use the chain rule. So the derivative of t ^2 is 2T hyperbolic tangent of 1 / t. Now the derivative of hyperbolic tangent is hyperbolic secant squared 1 / t. But then we have to take the derivative of that 1 / t and it's all going to get multiplied by the first term of t ^2. So the derivative of 1 / t is -1 / t ^2. The t ^2 here and the t ^2 here are going to cancel. So our final answer for our first derivative is going to be Y prime equal to T hyperbolic tangent of 1 / t minus hyperbolic secant squared of 1 / t. Thank you and have a wonderful day.