Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Derivatives of trigonometric functions. We're going to find the derivative of our six trig functions. First, we're going to show that the limit is H approaches 0 of cosine H -. 1 / H is going to really equal 0, so we're going to start by multiplying it by its conjugate cosine H + 1 over cosine H + 1. So we get the limit as H goes to 0 of cosine squared H - 1 / H cosine H + 1. Now by our Pythagorean identity, cosine squared H -, 1 is really just negative sine squared H Now we have an identity that says the limit as H goes to 0 of sine H / H is really one. So if we split this up, we could have sine H / H times negative sine H over cosine H + 1. So the limit as H goes to 0, sine H / H is one, but the limit as H goes to 0 of negative sine H over cosine H + 1. If we stick in zero, we know that the sine of 0 is 0 and the cosine of 0 is 1. So we end up with the opposite of 0 or -0 / 1 + 1, which is really zero and one times 00. We're going to want to keep that in our memory banks because we're going to use the identity of limit as H goes to 0, cosine H -, 1 / H equaling 0 a couple times. So if we want to find the derivative of sine X using the official definition, we're going to have sine limit as H goes to 0 of sine X + H minus sine X all over H. So the limit as H goes to 0 of using the expansion formula or the addition formula for our sine. So it's going to be sine X cosine H plus cosine X sine H minus the sine X all over H I'm going to then group together the terms that have the sine X in them. If I pull out the sine XI get left cosine H -, 1 all over H and I'm going to split this up into two separate limits. So I'm going to put a plus limit as H goes to 0 of cosine X sine H / H at the far end. Now we're going to split those up into. The limit is H goes to 0 sine X. The limit is H goes to 0 of cosine H -, 1 / H plus the limit is H goes to 0 of cosine X. The limit is H goes to 0 of sine H / H So we're going to end up with sine X * 0 plus cosine X * 1, or just plain old cosine X. So the derivative of sine is cosine. Let's look at the derivative of cosine using the same kind of idea. So if we look at the derivative of cosine, we know that the derivative of cosine by definition as the limit as H approaches 0 of cosine X + H minus cosine X all over H. If we do our expansion, we get cosine X, cosine H minus sine X sine H minus cosine X / H. That's one of our trig identities that we have to know. Cosine alpha plus beta is cosine alpha, cosine beta minus sine alpha, sine beta. So here we're going to group together the ones that have the cosine X and factor it out. So we have cosine X times the quantity cosine H -, 1 all over H. We're going to subtract the second term limit as H goes to 0 of sine X times the sine H / H So as H goes to 0, the limit of cosine X is just cosine X, but that cosine H -, 1 / H is 0. So then we're going to subtract. The limit is H goes to 0. Sine X is going to be sine X because there wasn't an H in it, and the limit is H goes to 0 of sine H / H is 1. So we end up with negative sine X. So the derivative of cosine X is negative sine X. These are ones we're going to need to have memorized. So the next one, let's look at the derivative of tangent. Let's think of tangent as sine divided by cosine. And then we're going to use our quotient rule. So if we thought about sine being our U value and our cosine being our V value, we know that the quotient rule says the derivative of the top times the bottom minus the derivative of the bottom times the top all over the bottom squared or U prime v -, v prime U / v ^2. So the derivative of the top we just proved the derivative of sine is cosine. So the derivative of the top cosine times the bottom which is also cosine minus. The derivative of the bottom derivative of cosine we just proved was negative sine times the top which is sine all over the bottom squared. The negative times negative makes it a positive and we end up with cosine squared X plus sine squared X all over cosine squared X. Pythagorean identity says cosine squared X plus sine squared X is really just one. So one over cosine squared X is secant squared X. So we get the derivative of tangent X equaling secant squared X. Similarly, you can find the derivative of cotangent by making it into cosine X over sine X. So if we thought about the derivative of the top negative sine X times the bottom minus the derivative of the bottom derivative of sine as cosine times the top all over the bottom squared. So in this case we get negative sine squared X, negative cosine squared X all over sine squared X. If we thought about factoring out a -1, we'd then end up with sine squared X plus cosine squared X, which we know is really just a trig identity for one. And so now we end up with -1 over sine squared X or negative cosecant squared X. So the derivative of cotangent X is negative cosecant squared X. So the next one derivative of secant. Let's do the quotient rule again using one over cosine. So the derivative of one is just 0. So the derivative of the top zero times the bottom cosine X minus the derivative of the bottom derivative of cosine is negative sine times the top one all over the bottom squared. So when we multiply and simplify, we get sine X over cosine squared X and we know that sine X over cosine X is really tangent. So we could think of that as sine X times cosine X divided by cosine X * 1 over cosine X or tangent X secon X. So the derivative of secon X is tangent X secon X. You can do the same thing for cosecant. So the six derivatives, the derivative of sine is cosine. Derivative of cosine is negative sine derivative of tangent as secant squared, the derivative of cotangent as negative cosecant squared, the derivative of secant as secant tangent the derivative of cosecant as negative cosecant cotangent. The way I personally help myself with remembering positives and negatives is the derivative of the ones. The trig functions that start with C, cosine, cotangent, cosecant all have negatives. So that's personally how I help myself remember. I'm not good at memorization, so if I can find patterns that helps me. So let's do an example. We have Y equal X ^2, cotangent X -, 1 / X ^2. We're going to use the product rule here. So the derivative of Y in terms of X, the derivative of the X ^2 times the cotangent X plus the derivative of the cotangent X times the X ^2, then we're going to subtract the derivative. We could think of 1 / X ^2 as X to the -2. So the derivative of X ^2 is just two X times the cotangent X. The derivative of cotangent X we just showed was negative cosecant squared X * X ^2 plus using the power rule, we're going to bring down the exponent of -2 and then we're going to take it to 1 less power, so plus 2X to the -3. Simplifying that up, 2X cotangent X -, X ^2 cosecant squared X + 2 / X ^3. Traditionally we write the trig functions at the end. If we have variables and coefficients, we put the trig at the end, so the two X cotangent X, the negative X ^2 cosecant squared X, those trig functions are coming at the end of the term. Another example, P equals sine Q plus cosine Q over cosine Q. So this is the quotient rule. So we're going to do the derivative of the top times the bottom minus the derivative of the bottom times the top all over the bottom squared. Well, the derivative of sine is cosine. The derivative of cosine is negative sine. The derivative of times the cosine Q plus the derivative of cosine was negative sine. So a negative and a negative made it a positive times the sine Q plus cosine Q all over cosine squared Q distributing out to get rid of the parentheses we get cosine squared Q minus cosine Q, sine Q plus sine squared Q plus cosine Q sine Q all over cosine squared. So the cosine Q sine Q's are going to cancel and cosine squared plus sine squared is one and one over cosine squared is just secant squared. Another way we could have done this problem if we look at the original. Another way to do it would be to think about sine over cosine being tangent and cosine over cosine being one. So the derivative of tangent was secant squared, and the derivative of 1 is a constant. Derivative of a constant is 0. So secant squared Q + 0 or just secant squared Q. There's more than one way to do math, and sometimes one method is easier than the others. It just depends on if you see it or not. Thank you and have a wonderful day.