 Hi everyone. I have here for you nine example problems of finding the derivative of a trigonometric function. I would encourage you after the first few examples to maybe pause this video as you watch it, try the problem first on your own, and then restart it to see how it works out and see if you did it correctly. So let's start with the first one. We have y equals sine of x over 2. Now this is one in which we're going to want to think of that two in the denominator as the coefficient one-half. So our derivative then, of course derivative of sine is cosine. The one-half remains as the coefficient, so we have one-half cosine of x, and that's it. The second one's equally straightforward. Derivative of x simply is one. Derivative of cosine is negative sine, pretty easy so far. Now this next example is one in which is going to require the product rule up on this front term. So let's keep that in mind as we work through it. So applying the product rule, and I'll put it in square brackets so we can offset it, we'd have 2x, we'll keep that first expression. Now we need to multiply by the derivative of cosine, which of course we know is negative sine. So that's the first part of the product rule. So then we keep the second term, so we keep cosine of x, multiplied by the derivative of 2x, so that of course is 2. Now in the back, after the minus sign, the two remains as the coefficient. Derivative of sine simply is cosine. So if we go ahead and simplify this, you'll notice that the two cosine of x's will cancel out, and all we have remaining, we can throw that negative to the front. We have negative 2x sine of x as our answer. So this one's pretty straightforward as we begin. Derivative of course of x is simply 1, derivative of tangent is secant squared. But hopefully now you're looking at this, thinking that it looks a little familiar to something, you can apply a Pythagorean identity here. Hopefully you recall the identity is 1 plus tangent squared of x is equal to secant squared of x, that's how the identity goes. So if we have 1 minus secant squared, that simply is going to be negative tangent squared of x. And that's something you'll want to look for on the trig problems is use of some of your identities. This next one's going to require product rule as well. So we'll keep the x multiplied by derivative of secant, which is secant tangent plus now we keep the secant and derivative of x of course is just 1. So we can just leave it like that. Now this is one of which we could factor out. We have a greatest common factor here in secant, so we could factor that out. And if we think about what we have remaining then, we have an x tangent of x, and then we need a 1. Remember you can always check these by doing distributive property to make sure you get the previous line back again. And that's how we would leave our answer. The next one will be a little bit similar to it, in that it's going to require product rule as well. Keep the x derivative of cosine is negative sine. And we keep the cosine derivative of x is simply 1. So there's not a whole lot we can do to this. We can throw that negative out in the front to make it look nice. Some people might even like to rearrange the terms. So it's cosine of x minus x sine of x, and that of course would be perfectly acceptable. So here's another one that's going to require product rule up here in the front, right up here. So once again I'll put that in square brackets just to offset it. So we shall keep the negative 3x. Derivative of sine is cosine. Now for the second part of the product rule, we will keep the sine multiplied by the derivative of negative 3x, which is negative 3. So there's the end of the product rule. So then we have the derivative of negative 5 cosine of x. Remember derivative of cosine is going to be negative sine. So that's going to make this a positive 5 now. So if we start simplifying, notice we'll be able to simplify these sine terms back here. So in the front we have negative 3x cosine of x. We can combine the sine terms to be positive 2 sine of x. Here we have another product rule. So we'll keep x squared. Derivative of secant is secant of x tangent of x. And we'll keep secant this time. Multiply by the derivative of x squared, which is 2x. So this is another one in which we could do a greatest common factor. We have a secant we can factor out as well as a single x, right? So our greatest common factor then is x secant of x. So now we just need to know what that's multiplied by. So think of it as what you're missing from the previous term if you think of the distributive property backwards. So we have x tangent of x and then plus 2. For our final example, this is a great example of one in which if you just take a couple moments to think about how you might be able to simplify it first, you make it a lot easier to find the derivative. Otherwise, working with it as it is, you'd have to use product rule. I think you know that. Having done it, it's pretty long and tedious. It's beautiful math though. It really is. But take a moment to think about how you might be able to simplify this. Remember, tangent is sine over cosine. Cosicant is a reciprocal of sine. So if you're a little bit clever in terms of how we might rewrite this, we would have sine over cosine and cosecant as 1 over sine. Notice how the sines cancel out and all we have is 1 over cosine, which is simply secant. So what we're really just doing the derivative of is secant of x and that's it. And you know that rule. Derivative of that, of course, is secant tangent. And you're finished. As I said otherwise, you would have had to do product rule, which gets pretty long. So this is why it's really necessary for you to know your trig identities well, so that when you come across problems like this, you get pretty savvy at how to rewrite them from the front end.