 Hello, and welcome to a screencast about derivatives of other trigonometric functions. So right now we know the derivatives of four trig functions, and there are two more, but those are left to the reader. So make sure that you do those exercises so you know what those derivatives are as well. So the derivative of sine of x is cosine. The derivative of cosine of x is negative sine of x. The derivative of tangent of x is secant squared of x. And the derivative of cotangent of x is negative cosecant squared of x. And I certainly would memorize the first two. The rest of them can be derived as we showed you in the book and probably in class. But if you want to memorize the rest of them too, that's perfectly fine. You know, it just depends on what makes you feel more comfortable. Okay, so for first example here, it says find the derivative of each of the following functions. And we have our function y equals e to the theta, tangent theta plus 3 cosine theta. So we have to kind of look here and I recognize a tangent theta and a cosine theta. So that's using the rules above. But I also see an e to the theta attached to this tangent of theta. So what rules are we going to need to throw at that? I see a 3 attached to this cosine of theta. What rules are we going to need to throw at that? So for this one, I mean you just got to kind of go through and build a pick out what pieces are going on here. But you notice we could say this first piece is e to the theta, tangent theta. Or we can also say e to the theta times tangent of theta. So because there's a little times idea in there, even though it looks like a decimal there, sorry. You know that you have to use the product rule, okay? So just depends on kind of how you want to organize this. But I'll say dy d theta because we're doing the derivative of y, that's the name of our function, with respect to theta. You could also call it y prime if you want to, but I just like to mix up the notation just to be good about those kind of things. All right, e to the theta times tangent theta. So because this is a product, we're going to need the product rule. So remember that one says you want to take the derivative of one function times the other plus the derivative of whichever one you didn't do times the other function, okay? So basically you're just kind of mixing up those derivatives and functions. So in this case, we're going to do the derivative of theta or with respect to theta of e to the theta times tangent theta. Then plus, so e to the theta is going to stay constant this time. The derivative with respect to theta of tangent theta. And then plus, we're going to have to do the derivative with respect to theta. Now, what do you guys think about this three times cosine theta? Because again, technically there's a time sign in there. Well, is three a function or is three a constant? Well, it's a constant, obviously, because it's just a number. So in this case, that can just come along for the ride like we saw before. So three cosine theta. If you really wanted to do the product rule on that one, you could, but the derivative of three is going to give you a zero. So that's just going to wipe out that piece of it. Okay, so I basically see one, two, three derivatives then that we need to do. Okay? So derivative of e to the theta, you should remember that. If not, go back and look for it. Derivative tangent theta, that's right up here above. And the derivative of three cosine theta, well, we can put our three together with the derivative cosine, which again is up above. So our derivative is going to be derivative of e to the theta is just e to the theta. Technically times the natural log of e, but remember that piece wipes out. So times tangent theta plus e to the theta times the derivative of tangent is secant squared of theta. I know it says x up here, but remember whenever your inside function, like the actual variable changes, you can make sure to be consistent with that. Plus then the derivative of theta of three cosine theta. Well, the derivative of three, or the three remember just comes along for the ride, derivative of cosine theta is negative sine theta. So I think I'm actually going to change this plus to a minus, just so it looks a little bit nicer, minus three sine theta. And that's your answer, okay. Next one I want to take a look at then, looks a little bit uglier. So we have v of t equals the square root of t sine of t, all over t cubed plus five cotangent of t minus one. Holy cow. Okay, so what do we have going on with this one? Well, overall I see a fraction. Okay, so fraction means we're going to use a quotient rule. And then within that, I also see a product in the numerator. Because again, I read this as square root of t sine t, but there's technically a multiplication in there. The square root of t times the sine of t. Okay, in the denominator, let's see, I see t cubed plus five, five times cotangent of t. Sorry, but five is a constant. So that's just going to come along for the ride. And minus one, we can deal with that piece too. Okay, so we're going to want to go back to the quotient rule. Let's recap what that one was. So if our function is written as f over g, which is how we'll just go with that, just so it's a little bit easier for us. That says we start with the denominator. So g times f prime, so the derivative of the numerator, minus the numerator times the derivative of the denominator. And then over the denominator squared. Okay, so, but like I said, within the numerator here, we're also going to need to do the product rule. Okay, so f prime is going to look like. So I'm going to, it doesn't matter again what order you want to do this in. So we'll say d dt of square root of t, but I'm going to take a second and go ahead and rewrite that as t to the one-half and then sine of t plus t to the one-half or square root, doesn't matter whichever way you want to write it. The derivative with respect to t of sine of t. Okay, so within that numerator, we're going to have to do two derivatives and then mix it all into the product rule. Okay, so let's go ahead and simplify this and then we'll get to our g prime in just a second. So the derivative with respect to t of t to the one-half, well, that's a power. So we can go ahead and use the power rule. So that's one-half t to the negative one-half, sine of t just comes along. And if you want to throw more parentheses in here, obviously go ahead. Plus, then we have t to the one-half derivative of sine of t is cosine of t. And sometimes I throw parentheses around here, sometimes I don't. As long as you remember that t is the argument of that function. You should be good. Okay, there really isn't much simplifying we can do with this. So let's just go ahead and leave it big and ugly like that. Okay, then now I'll come down here. g prime is going to look like, so the derivative of t cubed, that's just a power. So that's 3t squared plus the derivative of five cotangent of t. So if you recall, the derivative of cotangent is negative cosecant squared. So again, let me go ahead and change this plus to a minus, so that'll be minus. The five's a constant, so that just comes along for the ride. Cosecant squared of t, that extra parentheses came in there. And then minus one, the derivative of minus one is a constant. So that's just going to go away to zero. Okay, so now we have to take these four pieces that we just found and put it together in our quotient rule. So the overall derivative here is going to be, I'm going to keep switching colors on us. So v prime of t is the denominator. So that was t cubed plus five cotangent of t minus one times the derivative of the numerator. So that's this big junk that we got up here. One half t to the negative one half times the sine of t plus t to the one half times the cosine of t minus, okay, I'm going to run out of room here very quickly. F, so that was our original numerator. Square root of t or t to the one half, however you want to write it. Sine of t times the derivative of our denominator, which is three t squared minus five cosecant squared of t, got close there. And then all of this, whoa, is divided by our denominator, t cubed plus five cotangent of t minus one squared. That's a big one. All right, thank you for watching.