 In this video, I want to consider the derivatives of the other four trigonometric functions. We've already considered the derivative of sine and cosine. We know that the derivative of sine now is equal to cosine. We also know that the derivative of cosine is equal to negative sine of x. Using those two observations, we want to consider the derivatives of the other four trigonometric functions, tangent, cotangent, secant, and cosecant. You'll, and you see them here on the screen, we will see in just a second that the derivative of tangent is equal to secant squared. The derivative of cotangent is equal to negative cosecant squared. The derivative of secant is equal to secant times tangent, and the derivative of cosecant is negative cosecant times cotangent. Now that might seem like a lot to memorize, but let me give you a little bit of a pneumonic device to help you out here. So you'll notice that of the six trigonometric functions, they kind of come into two categories. There's those with the name co, that the prefix co is in their name, and those who don't have it, right? So you have the co, you these co-labeled functions like cosine, cotangent, and cosecant. The reason the co here is short for complementary, like complementary angles, these trigonometric functions come in complementary pairs, like secant and cosecant are complements of each other. Tangent and cotangent are complements of each other, likewise sine and cosine are complements of each other. If you compare the derivative of the trigonometric function without the co, and then you consider the derivative of the function with the co, you'll notice a very common pattern here, like notice tangent versus cotangent, right? The derivative of tangent is secant, great. The derivative of cotangent is a negative cosecant squared, right? So if you slap a co in front of tangent, you're going to slap a co in front of the secant, but you also get a negative sine, okay? Consider also the derivative of secant. Its derivative is secant times tangent, but what about the derivative of cosecant? It's a negative cosecant cotangent. So if you slap a co in front of secant, its derivative will be negative cosecant cotangent. You just put a co in front of each of the functions in an extra negative sine. That's how they're related. The same is also true for sine and cosine. If you take the derivative of sine, you get cosine. If you take the derivative of cosine, you get a negative sine. That is to say, the derivative of sine, if you slap a co in front of it, you're going to get a negative co cosine, right? That's kind of funny, right? You maybe want to drink some co cosine on a cold winter day, but if you take the complement of the complement, you get the original function, right? So here you see a negative sine, but if you think of it as the negative co cosine, then the pattern turns out to be true as well. The derivative of a trigonometric function if you know that you can then find the derivative of its complement by adding the negative sine, switching the sine, I should say, and switching to the co functions as well. So the reason I point that out is that if you know some of the trigonometric derivatives, you can actually find out the other ones. So in this video, we're going to talk about the derivatives of tangent. How do you prove the derivative of tangent is secant squared? And it turns out it's based upon the observation that tangent is equal to sine over cosine. That is, we can recognize that tangent is some type of ratio of sines and cosines, for which then if we take the derivative of sine divided by cosine using the quotient rule, we get low d high minus high d low square the bottom. Here we go. That is the derivative of sine over cosine will be cosine x times the derivative of sine of x minus sine of x times the derivative of cosine of x all over cosine squared of x. But as we observed over here, the derivative of sine is equal to cosine and the derivative of cosine is equal to negative sine, for which case we get a cosine cosine, which is a cosine squared. Then we're going to get a negative sine times a negative sine, which is actually a positive sine squared. And then we of course, arrive upon everyone's favorite trigonometric identity, cosine squared plus sine squared is equal to one the Pythagorean identity. So we can simplify the numerator to get one over cosine squared. The next thing to observe is that secant, of course, is just equal to one over cosine. So if we have one over cosine squared, that can be written as a secant squared. So that gives us the proof of the derivative of tangent. The derivative of tangent is equal to secant squared by taking the quotient rule since tangent sine over cosine. Well, cotangent is equal to cosine over sine. If you apply the quotient rule to that derivative using these little tidbits as well, you can show by a similar argument that cotangent's derivative is negative cosecant squared. Similarly, secant, of course, is one over cosine. You can use the quotient rule there as well and then cosecant is one over sine. So using the quotient rule will fill in the rest of these. I'll leave it up to the viewer here to verify those results as well. But you're going to want to know these, you're going to want to know these trigonometric statements right here. This is something worth memorizing so you don't have to re-derive them each and every time. Of course, if you had to, you can use the quotient rule to get the correct derivative. If you're in some type of desert island scenario playing crashes, only you and the volleyball survive. There's no graphing calculators around. But you have to calculate the derivative of secant to find home. Well, you can do that. Just use the quotient rule. No big deal. So let's take a look at such an example here. Let's find the derivative of f of x if it's equal to x squared times secant of x. We see this is a product of two functions, x squared and secant of x. So the derivative of f is going to look like by the product rule, x squared prime times secant of x. And then we're going to take the derivative of secant like so. So now the derivative of x squared by the power rule is going to be 2x. And then the derivative of secant, as we saw on the previous slide, that's going to equal secant of x times tangent of x. Again, that's just something you're just going to want to memorize, for which this gives us the correct derivative. As it is often useful to factor the derivative, notice there's a common factor of x and a common factor of secant. So if you wanted to, you could factor out x times secant of x. And that would leave then behind a 2 plus x tangent. That step's not really necessary here. But as we're going eventually to chapter 3, where factoring derivatives is going to be crucial, it's kind of a good practice to start doing it now. Because if we do it now, then we're going to be practicing it right now. When it's kind of low stakes, it's not necessary. So we just practice it now. Later on, it will be necessary. So let's get good at it before we get to that point. How about we want to find the derivative of f of x this time to be secant of x over 1 plus tangent of x. And let's use that derivative to figure out where the horizontal tangent lines are. So notice we're trying to find the tangent line of the tangent curve. Well, modification of the tangent curve. But hey, two different means of tangent in play right now. So to first calculate the derivative of f prime, we're going to use the quotient rule. So singing the song or just reciting out in your out loud, whatever you need to do, we're going to get low, d high, take the derivative of secant, minus high, d low, take the derivative of 1 plus tangent x, square the bottom, here we go. And then when you have that square in the denominator 1 plus tangent squared of x, again, there's no benefit to multiply out the denominator, leave it alone. And so we do have to calculate these derivatives. As we saw earlier, the derivative of secant, we just did that in the last example, that's going to be a secant of x times tangent of x, that's a product, minus secant of x. Now we have to take the derivative of 1 plus tangent of x, we'll take the derivative of 1. It says it's constant, the derivative will be 0, so it just disappears. We take the derivative of tangent, you're going to end up with a secant squared of x, again, this all sits above the 1 plus tangent of x squared. And so here we have our function, our derivative now calculated. We have to figure out the horizontal tangent lines, where are the horizontal tangent lines? Well, if this line is horizontal, that means that slope is equal to 0. But as it's a tangent line, the slope is actually the derivative. So we have to figure out where the derivative is equal to 0. So we want to solve this equation equal to 0. Now the good news is, whenever you have a fraction that's equal to 0, that only happens if the numerator is equal to 0. You can't make something equal to 0, the denominator is equal to 0. So basically times both sides by 1 plus tangent squared. And we end up with 1 plus tangent x times secant tangent minus secant cubed of x. This is equal to 0. So now that situation I was telling you about just on the last example, we've now arrived upon it. We need to factor this thing in order to figure out when it's equal to 0. You can factor out the common divisor of secant, that's really helpful. In which case that leaves you behind with a 1 plus tangent of x times a tangent of x. And then we have a minus secant squared, for which then I'm going to distribute this tangent through right here. Doing so is going to give us secant of x times we got tangent of x. We're going to get a plus tangent squared of x minus secant squared of x is equal to 0. Now some things to consider. Secant actually can never equal 0. Secant is 1 over cosine. And like I said a moment ago, the only way a fraction equals 0 is the numerator equals 0. If the numerator is 1, that's never going to be 0. So it turns out we can actually remove secant from consideration because it can't equal 0. My trigonometric identity sense is tingly and again tangent squared minus secant squared. It feels like there's a trigonometric identity somewhere in play right here. What can be done to help us with this calculation? Well you're going to want to remember the following trigonometric identity, right? 1 plus tangent squared is equal to secant squared. So notice if you subtract tangent from both sides, you're going to get 1 is equal to secant squared minus tangent squared. That's not quite what we have. The signs are opposites of times both sides by negative 1. Negative 1 is equal to negative secant squared plus tangent squared right here. So this friend is just equal to negative 1. You see right there. And so with that substitution, we see that tangent of x minus 1 is equal to 0. So solving for tangent, we get tangent of x is equal to 1. When is tangent equal to 1? Well it turns out that happens at pi force. But that's just in the principal region. It also would happen every time the tangent curve repeats itself, which is every 180 degrees or every pi radians. So if n is just any whole number, we see that the derivative will equal 0 whenever we are pi force plus a multiple of pi. And so this is where the horizontal tangents are going to occur on this graph. And we found this by computing and factoring the derivative.