 Hello, this is a video introducing you to the hyperbolic functions and Specifically we will discuss differentiating and integrating them first off the hyperbolic functions Whereas regular trig functions come from a unit circle these hyperbolic trig functions as they're known as Come from the unit hyperbola if you do want to figure out more information about this or would like to explore Feel free to look on the internet, but for the sake of this course It's just important that you know that they exist and that you know that they have derivatives and they have anti derivatives or integrals So this is the shape of hyperbolic sign hyperbolic cosine and hyperbolic tangent these shapes probably look familiar or look similar to other graphs that you've seen and algebra or pre-capitalist courses Then we have hyperbolic cotangent hyperbolic secant and we have hyperbolic cosecant so What's the purpose of these functions? Well one such purpose for instance could be a Cable suspending from two towers This cable that is being suspended from two towers actually takes the shape of the hyperbolic cosine function So that's one practical application Now the actual definitions or relationships for hyperbolic functions if you look at sign hyperbolic sign Hyperbolic sign by definition is actually e to the x minus e to the negative x over two hyperbolic cosine is e to the x plus e to the negative x over two and There's actual other definitions for the other hyperbolic functions in terms of the natural log But for the sake of this course Tangent for instance is still sine over cosine Cosecant still one over sine Secant one over cosine and cotangent one over tangent. Of course, that's all hyperbolic in those functions So understand that sine h of x or Sine hx is read as the hyperbolic sine of x. Some people simply just say cinch and for hyperbolic cosine Some people would just say coach Now as we work these examples Feel free to watch and listen first and then you can pause the video to write your work down. That's what you would prefer So the cool thing about hyperbolic functions is that they behave very similar to regular trig functions Not exactly the same but very similar So for instance if hyperbolic tangent is one half find hyperbolic cosine and find hyperbolic sine So we start with an identity and this can be found in your book at the very beginning of the section within the first page or two so one identity says that Hyperbolic tangent square root of x plus hyperbolic secant square root of x is equal to one Yes, this is one of the Pythagorean identities or very similar to one of the Pythagorean identities. I should say And so I have one half squared plus secant hyperbolic secant square root of x is equal to one When we square the half we do get a fourth When we take away a fourth from both sides, we are literally left with Hyperbolic secant square root of x equals three fourths and we'll square root of sides Leaving us with square root of three over two as hyperbolic secant now. You have to understand that Hyperbolic secant is the reciprocal of hyperbolic cosine. So if we flip hyperbolic secant That's how we get hyperbolic cosine And now to find hyperbolic sine To find hyperbolic sine we use the identity hyperbolic tangent is equal to hyperbolic sine over hyperbolic cosine So this means one half is equal to hyperbolic sine over two over the square root of three Since you're dividing by two over square root of three you would undo this by multiplying both sides by two over square root of three This leaves us with One over square root of three is equal to hyperbolic sine of x Part b if hyperbolic sine of x is equal to two-fifths. Let's find hyperbolic cosine and hyperbolic tangent So we start with An identity This is not the same as the Pythagorean identity. It's actually hyperbolic cosine minus hyperbolic sine is equal to one cosine squared sine squared I plug in two-fifths for hyperbolic sine So I get hyperbolic cosine squared of x minus two over five squared is equal to one So this turns out that being subtracting four over 25 from Hyperbolic cosine squared is equal to one After you add four over 25 to both sides You do get hyperbolic cosine squared of x is equal to 29 over 25 You square root both sides and you'll get square root of 29 Over five as hyperbolic cosine And now to find tangent. Well, we know that hyperbolic tangent is equal to hyperbolic sine over hyperbolic cosine We know that this is two-fifths stacked over square root of 29 over five And then we get hyperbolic tangent is two over square root of 29 I'm not really Too picky about whether you rationalize this or not. It's completely up to you as long as I know how you Know how you know how to use the hyperbolic functions to find other ones. You're good to go now The derivatives and integrals Remember I told you previously that the hyperbolic functions behave similar to the trig functions Not exactly the same but pretty similar So the derivative of sine is actually cosine the derivative of the hyperbolic cosine is actually just sine And this u prime here is just saying hey take the derivative of the inside of the function because of the chain rule And I'm not going to read through the whole list for you But some other things you can look at the antiderivative cosine hyperbolic cosine of u is equal to hyperbolic sine of u plus c The antiderivative of hyperbolic sine of u With respect to u is hyperbolic cosine of u plus c So you don't actually have that negative sign here like you would normally have for a regular trig function So that's one difference and then you have the rest of the antiderivatives I do not require that you memorize these which you should know how to use them So example two, let's differentiate y equals hyperbolic cosine of 8x plus one So to take the derivative we have y prime equals the derivative of hyperbolic cosine is Hyperbolic sine and then you keep your inside 8x plus one Because you had something other than x within the function It's your job by the chain rule to take the derivative of that inside block or that inside piece So this means that we have hyperbolic sine of 8x plus one Times eight that times eight does not multiply with the inside of the function It always goes out front of the function. So y prime equals eight hyperbolic sine of 8x plus one Now in part b You have natural log of hyperbolic tangent of x over two This is actually going to be a triple chain rule because nested within natural log is hyperbolic tan nested within hyperbolic tan is x over two Remember what the chain rule is like it's like peeling the layers of the onion You start with the outer most function first and work your way inside So we begin by taking the derivative of natural log So when we take the derivative of natural log, it's one over what's inside It's one over hyperbolic tan of x over two But then because you had something inside the natural log other than x you have to take the derivative of that inside piece You have to take the derivative of hyperbolic tangent of x over two So what this gives us is we still have our one over hyperbolic tangent of x over two, but now we have Hyperbolic secant squared of x over two But then since there's something inside the function other than x since there's something inside that hyperbolic function other than x We have to take the derivative of that inside. We have to take the derivative of the x over two Which does end up giving us just a half So at the end of the day when we put all this together we get hyperbolic secant squared of x over two over Two hyperbolic tangent of x over two So that's the final answer we will use But wait, there's a little bit more here part c f of x equals x squared times hyperbolic sine of x Plus hyperbolic cotangent of 6x So first to differentiate the x squared times hyperbolic sine of x we have to use the product rule We have to take the first factor, which is x squared and multiply it by the derivative of the second factor That's the derivative of hyperbolic sine of x So the first times the derivative of the second plus the second times the derivative of the first That's x squared times hyperbolic cosine of x plus hyperbolic sine of x times 2x So this gives us x squared hyperbolic cosine of x plus 2x hyperbolic sine of x To take the derivative of hyperbolic cotangent of 6x we will use the chain rule What is the derivative of hyperbolic cotangent? Well, it's negative hyperbolic cosecant squared Of 6x then take the derivative of that 6x afterwards So the chain rule does produce a factor of 6 remember that 6 does not multiply with the inside of the function It goes out front of the function. So we have negative 6 Hyperbolic cosine squared of 6x there should actually be an h here. It's there now Putting this all together gives me the derivative of x squared hyperbolic cosine of x Plus 2x hyperbolic sine of x minus 6 hyperbolic cosine squared of 6x So that's differentiating the hyperbolic functions well Without integrating So now we'll integrate and yes, we'll be using a lot of u substitution here So they give me the find the anti derivative or integrate hyperbolic cosine of 2x times hyperbolic sine squared of 2x with respect to x So i'm going to start off with The substitution let u equal hyperbolic sine of 2x The derivative of u with respect to x is hyperbolic cosine of 2x and then the chain rule produces an additional factor of 2 So after a little bit of rearrangement We're able to obtain that du over 2 is equal to hyperbolic cosine of 2x dx All right, so looking at our question here I have my hyperbolic cosine of 2x and dx Which we replaced with du over 2 and I have my Hyperbolic sine squared of 2x which I will replace that with u it will become u squared u squared du over 2 This means I have to have one half and then find the anti derivative or integrate u squared with respect to u Becomes u cubed over 3 the half that's still out front Replacing u back to being in terms of x. I have one over six two times three gives me six Hyperbolic sine cubed of 2x Plus c that's an indefinite integral. Therefore, we must have the plus c indefinite means I have no bounds of integration All right one last one So for u substitution, we would usually Set u equal to the inside of the function So let me just go out on them here and let u equal the square root of x, which is x to the half The derivative of u with respect to x is one half x to the negative one half So a little bit of rearranging we can move that x to the negative one half to the bottom of the fraction that becomes positive half And then this will become du equals one over two square root of x followed by dx Now I know that after rearranging two du is equal to one over square root of x dx. I like this I like the one over square root of x dx because Look at our Our integral here. I have my square root of x on the bottom and dx that can be replaced with 2 du In hyperbolic cosine. I have my square root of x that can be replaced simply with u All right, so looking at this I now have my integral rewritten as Hyperbolic cosine of u times 2 du That's two times the antiderivative or integral of hyperbolic cosine of u with respect to u That is using our integration formulas for hyperbolic functions 2 times hyperbolic sine of u Putting the integral back in terms of x we will get 2 times hyperbolic sine of square root of x that's what u is Plus c because it isn't it it is an indefinite integral So that's all I have for you now. I hope you enjoyed Thanks for watching