Welcome to Calculus One Jim Fowler, P5

Derivatives of Transcendental (Trigonometric) Functions

So far, we can differentiate polynomials, exponential functions, and logarithms. Let's learn how to differentiate trigonometric functions.

What are derivatives of transcendental functions?

We have learned how to differentiate just a ton of functions. If we can differentiate two functions, we can differentiate their sum. We can differentiate their difference. We can differentiate any polynomial. If we can differentiate two functions, we learned about the product and the quotient rule, so we can differentiate their product and their quotient. 

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We've also seen how to differentiate e to the x, and we know how to differentiate natural log. But there are still some functions that we can't differentiate. 

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We can't differentiate trig functions yet. We don't yet know how to differentiate sine or cosine, and that is now our task. What is the derivative of sine? What is the derivative of cosine? Once we know those derivatives, we will be able to calculate, very complicated expressions involving trig functions. Because we've already learned the change rule, which lets us differentiate the composition of functions. So if we can just figure out what the derivative of sine of x is, and what the derivative of cosine of x is, then we're going to be able to differentiate all kinds of very complicated expressions. 

What are transcendental functions?

, Welcome back to Calculus One. And welcome to week six of our time together. Can you believe that we've been at the Calculus game for more than a month already? And we've done a ton of stuff so far. Way back at the beginning of the course, we first met limits. We explored what it meant to evaluate a function near, but not at, a particular import point. And once we have limits, we were able to define the derivative. We explore how wiggling the input to a function would effect the function in some way. That ratio given by the derivative. Then, we spent a lot of time focusing on how to actually compute the derivative. Alright? We learned all these derivatives rules of the product rule, and quotient rule last with the chain rule for actually computing derivative of functions. And this week is no exception. We continue that proud tradition of actually differentiating. But in week six, we focus in on the transcendental functions. What are transcendental functions? Now, most of the functions that we've been looking at thus far are really algebraic functions. They're, say, polynomials or rational functions, roots. Transcendental functions are functions that transcend algebra, like e to the x, or log. Functions that we couldn't have gotten to if all we had at our disposal was algebra. Another great example of transcendental functions are sine, cosine, and tangent, the trigonometric functions. And if your not feeling super comfortable with trig functions, no worries. We're going to review trig functions this week, as well. But by the end of this week, we're going to know how to differentiate those trick functions, and that sets us up for week seven. And what happens in week seven? Applications of the derivative. We're spending a ton of time focusing on techniques of differentiation, and next week we'll see some applications. The pay off is going to be huge. Just one more week of calculations to get through, and then we can see what all of this is used 

Why does trigonometry work?

What's trigonometry? Well here's the basic idea. So I've got a bunch of right triangles, these are all right triangles, and this angle is the same in all of them, but they're all different sizes. So the side lengths are all different but the ratios between the corresponding sides are the same. What do I mean? Well, take a look at this big triangle. It's got some height that I'll label in, in orange, and it's got some width that I'll label in blue. Now the width of this triangle isn't the same as this triangle, isn't the same as this triangle, isn't the same as this small triangle. And the height of this triangle is different than the heights of these three triangles. But look at the ratio of this height to this width. Alright, here's the height, here's the width, and if I double the width, Width. It's a little bit more than the height. In other words the height is just a little bit less than the twice the width and that's true for all of these triangles. The ratio of their heights to their widths are all the same, even though they're different sizes. That's the key fact that makes trigonometry work. If you just draw some random right triangle with given angles, it doesn't really make a lot of sense to ask questions about the side lengths, it really depends on how big the triangle you drew. But it does make sense to ask questions about the ratios of side lengths, those only depend on the angles. So we can give names to these ratios. We call the sine of theta as the ratio between the height and the hypotenuse. So if you like, the opposite side and hypotenuse. So y over r in this diagram. The cosine of theta is the ratio between the adjacent side and the hypotenuse of the right triangle. It's x over r in this picture. And the tangent of theta. That's the opposite side over the adjacent side, it's y over x in this picture because the sine, cosine, and tangent of this angle theta. And they're all just defined in terms of ratios of side lengths. Now that we've got a definition of sine. We can, for instance, calculate sine of pi over 5. If we build a triangle, any right triangle with an angle of measure pi over 5. I built a right triangle. And this angle is pi over 5. I can use my little ruler to measure the hypotenuse. The hypotenuse here is about 25.2 centimeters. And I can measure the opposite side, and the opposite side's about 14.8 centimetres. And sine is the opposite side over the hypotenuse, which means sine of pi over five is about 14.8. Over 25.2, which is about 0.59. 

Why are there these other trigonometric functions?

So we know about sine and cosine but what about all those other funny trig functions? So maybe we already know about sine and cosine, we saw tangent already too. But tangent could be defined in terms of sine and cosine as sine theta over cosine theta. Cosecant is just defined to be 1 over sine theta. Secant is 1 over cosine theta and cotangent is 1 over tangent theta. What's so special about those 6 functions? I mean why we'd have a function called cosecant if its just 1 over sine theta. Why don't we always just try 1 over sine theta. For that matter why even have a function called tangent if you can compute tangent in terms of sines and cosines. It doesn't seeming like we need all of these functions. To make matters worse, there's even other functions that practically no one knows about anymore. In addition to these, there's the haversine function which is just defined to be sine squared of the angle over 2. I mean you really don't need haversine once you've got sine. Considering that most of these functions can be defined in terms of other ones, the reason for studying, you know, these 6 trig functions isn't really mathematical. You don't really need cosecant if you've got sine. The reason for emphasizing these 6 trig functions is cultural. These are the functions that people are likely to see when they open some technical manual and knowing about these functions can sometimes give you the right intuition for how to attack a problem. Because sine and cosine are the legs of a right triangle with angle theta and hypotenuse length 1 just by the Pythagorean theorem we get this that sine squared plus cosine squared is equal to 1. Now if you believe this identity and you could divide this identity by cosine squared you get sine squared over cosine squared plus cosine squared over cosine squared is 1 over cosine squared. Now because we've got all these other functions I could rewrite sine squared over cosine squared as tangent squared, cosine over cosine is 1, and 1 over cosine squared is secant squared. And this is maybe an example of how knowing about the other functions can be helpful. Its a little bit easier to internalize this identity, I mean if you walk around and you happen to notice that secant squared you can think always going to place over tangent squared plus 1. Then say trying to internalize this middle identity. This also a lovely geometric picture that kind a sells you on an idea that these 6 trig functions have some special role. For example, here is a unit circle and I've drawn an angle of measure theta and we know what some of the lengths in this picture are. This length across the bottom here is cosine theta. And this length here, the height of that right triangle is sine theta. But it turns out that the other 4 trig functions are also encoded in the lengths of other relevant lines in this diagram. For instance this line here has length tangent theta and this line between the point and the y axis has length cotangent theta. And if you measure on the bottom here the length from the origin to where this tangent line crosses that has length secant theta. And, if we measure over here, from the origin to where that tangent line crosses the y axis, that segment has length cosecant theta. So when you put all these functions down we've got cosine and sine but we've also got very visibly cotangent, tangent, secant, and cosecant. And the secant is measuring you know something that's crossing the circle and the tangent is really measuring the length of part of this tangent line. And if we take this picture and we go fader you won't get idea of how all the trig functions vary together. For instance by, by looking at this picture with theta moving you can get a sense of how these functions are moving together. For instance, tangent and secant are moving together right when tangent is really big, secant is really big. And conversely when cotangent is really small, cosecant is you know small, close to 1. They can use some idea as to why tangent is called tangent. Why is sine called sine? So sine comes from this Latin word sinus, you know, like the thing in your nose. It's like a opening, and if like that as an device you can remember that sine is measuring this side of the right triangle because you can imagine the right triangle sort of opens up in that side. 

What is the derivative of sine and cosine?

. Once, we believe that sine and cosine are important functions, they're all about the connection between angles and lengths. Well, then we want to apply our usual calculus trick. What happens if I wiggle the input to sine and cosine? You might think about trigonometry as being something about right triangles. But, here, I've drawn a picture of a circle and you can rephrase all this stuff in terms of just geometry of the unit circle. So this is the unit circle. The length of this radius is 1 and I've got a right triangle here and this angle is theta. That means this base here has like cosine theta and the height of this triangle is sine theta. The coordinates then, or this point on the unit circle are cosine theta sine theta. Now, let's make that angle just a little bit bigger. Let's suppose that I make this angle just a big bigger, and instead of thinking about theta, I'm thinking about an angle of measure theta plus h and that means this angle in between has measure h. We'll do a little bit of geometry to figure out how far that point moved when I wiggled from theta to theta plus h. So let's think about this point here and let's draw the tangent line to the circle at that particular point. Well, I'm going to draw a little tiny triangle, which is heading in the direction of that of that tangent line. And, so that it's tangent to the circle, the hypotenuse of that little tiny right triangle here in red will be perpendicular to this line here. I can actually determine the angles in that little, tiny right triangle. So we go this big right triangle down here, and just because its a triangle, this angle theta plus this angle in here plus this right angle have to add up to 180 degrees. But I've also got a straight line here and that means that this top angle plus this right angle plus this mystery angle must also add up to 180 degrees. Well, this plus this plus this is 180 degrees. And this plus this, same right angle, plus this mystery angle add up to 180 degrees. That means that, that mystery angle and that little tiny right triangle must also be theta. I'd also like to know the length of the hypotenuse of the little tiny triangle. Radians save the day. How so? Well, I want to know the length of this little piece of arc. What else do I know? I know this is a unit circle. And I know this angle here is h in radians. And the definition of radiance means that this little length of arc here has length h. Now, let's put my tiny right triangle back there. I'm going to have the hypotenuse of that little right triangle also be h. It's going to be close enough, right, because this little piece of curved arch and this straight line are awfully close. Now that I know the length of the hypotenuse and the angles in that right triangle I can use sine and cosine to determine the side lengths of the other two sides. So I've got a little tiny right triangle, hypotenuse h, that angle there is theta, and that means that this vertical distance here is h times cosine theta. And the horizontal distance of that little tiny red triangle is h sine theta. Okay. So how much did wiggling from theta to theta plus h move the point around the circle? So the original point here from angle theta had coordinates cosine theta sine theta. And the point up here which I got when I wiggled theta up to theta plus h, that point has coordinates cosine theta plus h sine theta plus h. So how much did wiggling from theta to theta plus h move the point? Well, if I use this little right triangle as the approximation, sine theta increased by about H cosign data, and cosign data decreased by about h sine theta. We're now in a position to make a claim about the derivative. In other words, from the picture, we learned that sine theta plus h is about sine theta plus h cosine theta. And cosine theta plus h is about cosine theta minus h sine theta. And as a result, we can say something now about the derivatives. How does changing theta affect sine? Well, about a factor of cosine theta compared to the input. So the derivative of sine is cosine theta. And how does changing theta affect cosine? Well, about a factor of negative sine. So the derivative of cosine is negative sine. Maybe you don't find all this geometry convincing. Well, we could go back to the definition of derivative in terms of limits and calculate the derivative of sin directly. So if I want to calculate the derivative of sin using the limit definition of the derivative, well, the derivative of sine would be the limit as h approaches 0 of sine theta plus h minus sine theta over h. The trouble now, is that I've gotta somehow calculate sine theta plus h. How can I do that? Well, you might remember, there's an angle sum formula for sine. Sine of alpha plus beta is sine alpha cosine beta plus cosine alpha sine beta. If I use this, but replace alpha by theta and beta by h, I get this. Sine of theta plus h can be replaced by sine theta cosine h plus cosine theta sine h. So let's do that. So, the derivative is the limit as h approaches zero of, instead of sine theta plus h, it's sine theta times cosine h. This first term, plus cosine theta sine h minus sine theta from up here and this whole thing is divided by h. So, this is sine theta plus h minus sine theta all over h. Now, I can simplify this a bit. I've got a common factor of sine theta, so I can pull that out. This is the limit as h approaches zero, pull out that common factor of sine theta. What's left over is cosine h minus 1 over h plus, I've still got this term here, cosine theta sine h over h. I'll write that as cosine theta times sine h over h. All right. So this limit calculates the derivative of sine. Now, it's written as a limit of a sum, which is a sum of the limits, provided the limits exist. So, I can also note that sine theta and cosine theta are constants. Right, h is the thing that's wiggling, so I can pull those constants out of the limits as well. So what I'm left with is sine theta. Times the limit as h approaches 0 of cosine h minus 1 over h. Plus cosine theta times the limit as h approaches 0 of sine h over h. Now, how do I calculate these limits? Well, we've got to remember way back to when we were calculating limits a long, long time ago. We can calculate these limits by hand, using say, the squeeze theorem. And it happens that this first limit is equal to 0 and the second limit is equal to 1. So I'm left with 0 plus cosine theta. So, this is a limit argument, back from the definition of derivative that the derivative of sine is cosine. So, regardless of whether you think more geometrically or more algebraically, the derivative of sine is cosine and the derivative of cosine is minus sine. Now, once you believe this, there is some sort of weird things you might notice, like what if you differentiate sine a whole bunch of times? So the derivative of sine is cosine and the second derivative of sine, is the derivative of the derivative. It's the derivative of cosine which is minus sine. And the third derivative of sine is the derivative of the second derivative and the derivative of minus sine is minus cosine. And the fourth derivative of sine, well, that's the derivative of the third derivative, which is minus cosine. And the derivative of a cosine is minus sine. So, the derivative of minus cosine is sine. So the fourth derivative of sine is sine. So that's kind of interesting. If you differentiate sine four times, you get back to itself. And we already know a function that if you differentiate just once, it spits out itself again, e to the x is its own derivative. So as sort of a fun challenge, you might try to find a function f, so that if you differentiate it twice, you get back the original function, but if you differentiate it only once, you don't get back the original function. And if you can do this, you can ask the same question for even higher derivatives. We know a function whose fourth derivative is itself, but none of the earlier derivatives are the function again. Can you find a function whose third derivative is itself, but the first and second derivatives aren't the original function? That's a fun little game to play, anyway, it's a little challenge for you to try to find such a function.

What is the derivative of tan x?

I want to differentiate tangent theta. How am I going to do this?  Well, hopefully you remember that tangent theta is sine theta over cosine theta.  Why is this? If you think back to the business about the right triangles, right, sine is this opposite side that I'm calling y over the  hypotenuse. And cosine is this adjacent side whose  length I'll call x over the hypotenuse. I'll call this angle theta. Then sine theta is y over r, and cosine theta is x over r, so this fraction can  be simplified to y over x. That's the opposite side over the  adjacent side. That's tangent of theta.  So good. I can express Tangent theta as sin theta  over cosine theta, how does that help? Well, I know how to differentiate sine  and cosine, and by writing tangent this way, tangents now a quotient, so I can  use the. Quotient rule.  All right, so I'll use the quotient rule, and the derivative of tangent theta is  the derivative of the numerator, which is cosine, times the der, denominator, which  is just cosine, minus the derivative of the denominator, which is minus sine,  times the numerator, which is sine. And this whole thing is over the  denominator squared, cosine squared theta.  Now, is this very helpful? Well, I've got cosine times cosine, so I  can write that as cosine squared theta. And here I've got minus negative sine  theta times sine theta, so that ends up being plus sine squared theta.  And the whole thing is still being divided by cosine squared theta.  Can I simplify that at all? Well cosine squared plus sine squared,  that's the Pythagorean identity. That's just one.  So I can replace the numerator with just one, still over cosine squared theta.  And one over cosine squared theta, well, one over cosine is called secant, so this  is really secant squared theta. .  So what all this shows is the derivative of tangent theta is second squared theta.  A moment ago we did a calculation using the quotient rule to see that the  derivative of tangent theta is second squared theta.  And now I want to see how this plays out in some concrete example to get a, a real  sense as to why a formula like this is true.  So here's a couple triangles. They're both right triangles and the  length of their hypotenuse is the same. This angle I am calling alpha, let's this  be a little bit less than 45 degree. And this angle I am calling beta, and  it's more than 45 degrees. Now what you can say about the Secant of  alpha and the Secant of beta? You know, the Secant of alpha is  definitely bigger than one. I mean, what's the Secant?  It's this hypotenuse length divided by this length here.  And that's bigger than one. How does it compare to beta?  Well, the secant of beta is quite a bit bigger than the secant of alpha, and why  is that? Well, the secant of beta is this length  here, the hypotenuse, divided by this width, but this triangle is quite a bit  narrower than this triangle. So the secant has the same numerator, the hypotenuse, the same length, but the denominator here is quite a bit smaller,  and if the denominator's a lot smaller, then the ratio, which is the secant, is  quite a bit bigger. Some of the secan of beta is bigger than  the secan of alpha, and the secan of alpha is bigger than one.  And that means that secan squared alpha is also bigger than one.  And secan squared alpha is less than secan squared beta.  And the significance of that is right here, the derivative of tangent theta is secan squared theta. So what does this mean?  Well, this, this is telling me how wiggling theta affects tangent.  It affects it like a factor of secant squared theta.  So in this example, where secant squared beta is a lot bigger than secant squared  alpha, the effect of wiggling beta on the tangent of beta should be a lot larger  than the effect of wiggling alpha on the tangent of alpha.  And you can see that. Let me draw a triangle, where I've  wiggled the angle alpha up a little bit. I've made it a little bit bigger.  But I'm going to make the same hypotenuse.  All right, so this hypotenuse length is the same as this length, but I've made  the angle a little bit bigger. And how is the tangent of the slightly larger alpha compared to the tangent of alpha.  Well, the tangent of the slightly larger alpha is bigger than tangent alpha, but  not by all that much. Now compare that to when I wiggle beta up  by the same amount. I make beta a little bit bigger.  Right. So I make beta a little bit bigger by the  same amount that I made alpha larger. And I think about how that affects the  tangent of beta. The tangent of beta is this height  divided by, divided by this width. And you can think about it, I mean this,  this height maybe isn't increasing a whole lot.  But the width of this triangle is getting quite a bit narrower.  And because it's the ratio of that height to that width the tangent of the  perturbed value of beta is quite a bit larger than the tangent of beta.  And you can, you know it's reflected in the fact the secant squared beta is a lot  larger than secant squared alpha. So these, these kind of facts, right?  The fact that the derivative of tangent is secant squared theta you, you can  really get a sense for why these things might be true by thinking about triangles  and how wiggling the angle will affect certain ratios of sides of the triangles.  But, if this seems a little too abstract we can kind of pull back a little bit and  do do a numeric example next. You know, and maybe the numerical example  is sort of another way to see a formula like this in action.  Let's do a numerical example to get a sense as to what you might do with the  fact that the derivative of tangent is secen squared. Here's an example, let's try to approximate the value of tangent of 46  degrees. Why is this an interesting example?  Well, we know the tangent of 45 degrees exactly, all right.  And figure that out by looking at a triangle, here's the angle, 45 degrees, a  right triangle, because the an, angles add up to, 180 degrees.  So it's 45 plus 90 plus what? Well, this must also be 45.  It's an isosceles triangle now, so these two sides are the same.  A tangent is the ratio of this side to this side because they are equal that  ratio was one. So I know the tangent of 45 exactly.  It's one. But I am trying to figure out an  approximation for the tangent of 46 degrees, the derivative tells me how  wrigly an input affects the output, so I can use this fact and the fact that I  know the derivative to try to approximate the tangent of 46.  In particular the tangent of 46 degrees is the tangent of 45 plus one degree.  And here you can see how I am perturbing the input a bit.  And this is exactly what the derivative would tell me something about.  A little bit of bad news, the derivative of tangent is only secant squared if I do  the measurement in radians. If I convert this to a problem in  radians. With radians, says the tangent of pi/4,  which is 45 degrees, plus with one degree in radians, is pi/180 radians, right?  This is what I want to compute. I want to compute the tangent of pi/4  plus pi/180. And I'll do that with approximation using  the derivative. So according to the derivative this is  about the tangent of pi over four, which is the tangent of 45 degrees, it's one.  Plus the derivative of pi over four, which is secant squared pi over four.  Times how much I wiggled the input by, which is pi over 180.  I know the tangent of pi over four. It's one.  What's the secant of pi over four squared?  Well, if I pretend that these sides have length one, by the pythagorean theorem,  this side must have length square root of two.  The secant is hypotenuse over this width. So the secant of pi over four is square  root of two. So secant squared pi over four is square  root of two squared, which is two times pi over 180.  Now if you know an approximation for pi, you can compute two times pi over 180  plus one. And this is approximately 1.0349,  and it keeps going Pi's irrational. And this is not so far off of the actual  value. If we actually compute tangent of 46, the  actual value is about 1.0355 and it keeps going.  And¡ 1 0355. is awfully close to 1.0349. So we've  successfully used the derivative to approximate the value of tangent 46  degrees.  

What is the derivative of sin(x^2)?

. Let's combine the change rule and the derivative of sign to differentiate a slightly more complicated function. Let's try to differentiate sine of x squared. We can realize this function as the composition of two functions. I can write sine of x squared as the composition of f and g, where f is the sine function and g is the squaring function. Now, how do I differentiate a composition of two functions? I use the chain rule. So I differentiate f and the derivative of f is cosine x, where the derivative of sine is cosine. And the derivative of g is just 2x. So now I want to differentiate the composition of f and g and that's by the chain rule f prime of g of x times g prime of x. In this case, f prime is cosine, so it's cosine of just g of x, which is x squared times the derivative of g, which is 2 x. So, the derivative of sine x squared with respect to x, is cosine of x squared times 2x. Honestly, this is a pretty neat example. In magnitude, this function, sine of x squared, is no bigger than 1. And yet, what do we know about this function's derivative? Well, the derivative of sine of x squared is cosine of x squared times 2x. And that function can be as large as you like. You can make cosine of x squared times 2x as big as you want, as long as you choose x appropriately. So what we have here is a function which isn't very big. The function's value is no bigger than 1 in magnitude, but the function's derivative is very large. And you can see that on the graph. The values of this function really aren't that large. The values are all hugging zero. But the derivative, the slope of the tangent line is enormous. Look over here. If you imagine a tangent line, that tangent line is going to have enormous slope. The derivative over here is going to be very large. In spite of the fact, that the actual values of the function really aren't that large. This actually provides another lesson. Just because 2 functions are nearby in value, doesn't mean that their derivatives are anything close to each other. For instance, here is the graph of the cosine function. And here is the graph of a function sine of x square over 10 plus cosine of x. Since sine of x squared is between minus 1 and 1, this differs from the cosine by no more than a tenth. And, yeah, you can see the graph is really close to the graph of cosine and yet this graph is way more wriggly. Let's zoom in and we can see the same sort of thing. Here's a zoomed in copy of just a cosine curve. And here's what happens if you zoom in on this other function. And in terms of the value, this other function really isn't different from cosine very much. But in terms of derivative, this function is totally different than cosine. This function is super wiggly, so the derivative of this function is enormous, even though the derivative of cosine is no bigger than 1 in magnitude. If you think this is kind of an interesting example, it's worth trying to cook up an even more elaborate example. Here's a very specific challenge for you. Can you find a function that, just make one up, so that your functions values and magnitude are less than c, and your functions derivative and magnitude is less than c? So I want a specific number c, so that no matter what value of x you plug in, the function's value there and the function's derivative there is less than c in magnitude. But, I want that function to have second derivative which can't be bounded by a constant. I want you to cook up a function so that, yeah, the values of the first derivative are bounded by c. But the second derivative can be as big or as negative as I'd like, by choosing x appropriately. Can you find a function like that? 

What are the derivatives of the other trigonometric functions?

We've seen how to differentiate sine, and cosine and tangent, but how do I differentiate secant? So let's differentiate secant. And one way to get a handle on secant is to write secant as a composition of two different functions. If I set f of x equals 1 over x, and g of x equals cosine x. Then f of g of x is secant x. So if I want to differentiate secant, it'd be good enough to differentiate f of g of x. And I can do that differentiation problem by using the chain rule. Now I can separately calculate the derivative of 1 over, it's negative 1 over x squared. And the derivative of cosine is negative sine. So putting this together, the derivative of secant must be the derivative of f, which is minus 1 over its input squared at g of x and g is cosine. So, with negative 1 over cosine square that's f prime of g of x times the derivative of g which is minus sine. Now I can put this together, the minus signs cancel and I'm left with sine x over cosine squared x. But people don't usually write it this way I could instead write this, as sine x over cosine x times 1 over cosine x. In other words, I could write this as, what's this term, this is tangent and this is secant so I can write this as tangent x secant x. So the derivative of secant x is tangent x times secant x. We can play the same kind of game to differentiate cosecant. So what about the derivative of cosecant x? Well think about how I can rewrite cosecant, alright? Cosecant is the composition of the one over function and sine, right? Cosecant is one over sine. So this is the derivative of f of g of x, where f is 1 over and g is sine, and by the chain rule, that's the derivative of f at g times the derivative of g. And now I know what the derivatives of these functions are. The derivative of 1 over is minus 1 over x squared, and the derivative of g is cosine x. And consequently, the derivative of cosecant x must be minus 1 over, that's the derivative of f at g of x which is sine, so it's sine squared times the derivative of g, which is cosine x. And I can combine these to get minus cosine x over sine x times 1 over sine x. In other words, minus cosine over sine, that's minus cotangent x, and minus 1 over sine, well that's cosecant again. So the derivative of cosecant x is minus cotangent x times cosecant x. And we can complete the story by differentiating cotangent. Okay, so how do I differentiate cotangent x ? Well we actually have some choices. One way would be to write this as the derivative of cosine x over sine x, since cotangent is cosine over sine. And then about the quotient rule. Or, I could use the fact that cotangent is one over tangent, and then differentiate this using the chain rule. To differentiate 1 over something, that's negative 1 over the thing squared times the derivative of the inside function which is tangent x. And I know the derivative of tangent x that's secant squared, so this is negative 1 over tangent squared x times secant squared x and that means the derivative of cotangent is negative secant squared x over. Tangent squared x. But people don't usually write it like this. Instead, I could simplify this. Having a secant squared in the numerator is as good as having a cosine squared in the denominator, and having a tangent in the denominator, Is as good as having cosines, in the numerator, and sines in the denominator. So, a negative secant squared over tangent squared is the same as negative cosine squared over cosine squared times sine squared. Now, these cosines cancel And what I'm left with is negative 1 over sine squared x. But people don't even write it this way, right? 1 over sine, well that's cosecant so this is actually negative cosecant squared x. Now that we've seen the derivatives of all six of our trig functions, how are we suppose to remember these derivatives? Well here's a table. Showing all the derivatives of these 6 trig functions. And you can see there's some real pattern to this table. The derivative of sine is cosine. The derivative of cosine has a negative sine. The derivative of tangent we saw is secant squared. And the derivative of cotangent is cosecant but with a negative sine. And the derivative of secant we just saw is secant tangent. And the derivative of cosecant is negative cosecant cotangent. So there's definitely some symmetry here, and you can exploit that symmetry to help you to remember these derivatives. 

What are inverse trigonometric functions?

>> People often talk about inverse trig functions. But that's nonsense. The trig functions aren't invertible. Look at y equals sine x. Sine sends different input values, like 0, pi, 2 pi, 3 pi, to the same output value of 0. So, how are you going to pick the inverse for 0? To get around this problem, we're only going to talk about the inverses of trig functions after we restrict their domain. Here I've got a graph of sine. And you can see this function is not invertible. It fails the horizontal line test. But, if I restrict the domain of sine to just be between minus pi over 2, and pie over 2, this function is invertible. And, here's its inverse, arc sine. By convention, arc sine outputs angles between minus pi over 2 and pi over 2. And also by convention, arc cosine outputs angles between 0 and pi. And arc tan outputs angles between minus pi over 2 and pi over 2. Note that I'm calling these inverse trig functions arc whatever. You know, arc cosine, arc sine. One reason to call these things arc cosine or arc whatever, is because of radian measure. If this is a unit circle, then the length of this arc is the same as the measure of this angle, in radians. That's the definition of radians. So, to say that theta is arc cosine 1 half, is just to say that theta is the length of the arc whose cosine is 1 half. Now, what happens when we compose the inverse trig functions with trig functions? Before complicating matters by thinking about trig functions, let's think back to an easier example, the square root and the squaring function. And just like trig functions are not actually invertible, the squaring function's not invertible, because multiple inputs yield the same output. So we had to define the square root to pick the non-negative square root of x. And that means that if x is bigger than or equal to zero, then I have the square root of x squared is equal to x. But if x is negative, this is not true. The same sort of deal happens with trig functions. So if theta is between minus pi over 2 and pi over 2, then acrc sine of sin of theta is equal to theta. But if theta's outside of this range, this is not the case. If theta's ouside of this range, yes, arc sine is going to produce for me, another angle, whose sine is the same as the angle theta. But there's plenty of other inputs to sine which yield the same output. Things are even more complicated if you mix together different kinds of trig functions. For example, the sine of arc tan of x happens to be equal to x divided by the square root of 1 plus x squared. Where would you possibly get a formula like that? Well the trick is to draw a right triangle, the correct triangle. Well here I've drawn a right triangle. And I've got an angle theta. And I've drawn this triangle so that theta's tangent is x. That means that theta is the arc tangent of x. Now by the Pythagorean theorem, I know the hypotenuse of this triangle has length of square root of 1 plus x squared. And that gives me enough information to compute the sine of theta. The sine of theta is this opposite side x, divided by the hypotenuse, the square root of 1 plus x squared. This tells me that the sine of arc tan of x must be x over the square root of 1 plus x squared. Drawing pictures will get you very far in understanding all of these relationships. I think it's basically impossible just to memorize all of the possible formulas that relate inverse trig and trig functions. But if you're ever wondering what those formulas are, you just draw the appropriate picture, and then you can figure it out right away. 

What are the derivatives of inverse trig functions?

What's the derivative of arc sine?  Well, I can't really compute this directly.  I don't know how to differentiate arc sine right off the bat.  So, I'm going to try to sneak up on the derivative of arc sine. Now, instead of writing down arc sine all the time, I'm going to give it a  different name. I'll write f of x for arc sine x.  It just, it looks shorter. Now, what do I know about arc sine?  Arc sine is the inverse function for sine.  So, at least for a range of values of x, f of sine x is x,  right? Arc sine tells me a value that I can take  the sine of to get the input to arc sine, so if I evaluate arc sine at sine x, I  get x, at least over a particular range of values of x.  So if this is true, and if f were differentiable, I could apply the chain rule and differentiate both sides.  So, let's just do that. I'm just plowing ahead assuming that f is  differentiable. So, if I assume that f is differentiable,  I differentiate f. I get some mystery derivative.  I don't know what it is yet, I'm calling it f prime.  At the inside, which is sine x, times the derivative of the inside, which is cosine  x. And this is equal to the derivative of  the other side which is one. So, what I know now is that the  derivative of arc sine at sine x is one over cosine x.  Try to divide both sides by cosine. But, what I really want is a formula that  tells me the derivative of arc sine at some point is some other point, alright?  I don't want to know the derivative at sine of x, in terms of cosine.  But, I can use a trick, right? sine squared plus cosine squared is one.  So if I knew that cosign were positive, I could rewrite this and you should be a  little skeptical as the quality, is one over the square root of one minus sine  squared x. You have to be a little bit worried here  because you got to know that cosine's positive.  Okay. But let's suppose it is. So, if cosine's positive, then I can write cosine x as the square root of one minus sine squared  x, and now I've got a formula for the derivative of arc tan, derivative of f,  at sine x is something involving sine x. So, I could rewrite this formula as just  the derivative of arc sine at x, is one over the square root of one minus x  squared. So, there we go.  There is a formula for arc sine assuming that arc sine is differentiable.  Alright? Because what I did is I just wrote down a  function, you know, that I knew to be true.  I know that fsinx sine x is x for certain values of x.  And I just plowed through, differentiating it, assuming that it was  differentiable. And then, I figured out what a derivative would have to be if it were differentiable.  But note that I never actually verified that arc sine is differentiable.  I'm just telling you what the derivative is if it were differentiable.  So that's perhaps a little bit unfortunate.  Alright. But nevertheless we're going to use this  and this is in fact, the derivative of arc sine.  Now, we know how to differentiate arc sine.  Let's try to differentiate arc cosine. We could do that in the same way.  f equals arc cosine to the chain rule trick.  We get a formula for the derivative of arc cosine.  Assuming it's differentiable, and it is. But, I want to do it a different way.  I want to try to do it different attack, a different approach on the problem of calculating the derivative of arc cosine. So, here's how we're going to start.  Take a look at this triangle here. this angle's alpha, and I'm labeling this  side to have length y, and the hypotenuse of this right triangle  has length one. Now, what this triangle's telling me is  that the sine of alpha, which is this sign divided by hypotenuse, is y.  And consequently, the arc sine of y is alpha.  Now, what's arc cosine of y? To figure that out let's draw another one  of these triangles. I'm going to label this side beta, and  I'm going to imagine picking up this right triangle and flipping it over.  So that then this angle's beta, this angle's alpha, hypotenuse is still length  one, but the side that I labeled y is now down here.  And what this triangle is telling me is that the cosine of beta, right? Which is  this width divided by the length hypotenuse is is y.  And consequently, the arc cosine of y is beta.  Why is that significant? Well, this is a right triangle and the  angles add up to 180 degrees, so alpha plus beta add up to 90 degrees, or pi  over two radiants. So, that means what?  I know that alpha plus beta is equal to pi over two.  Consequently, arc sine y plus arc cosine y, which is alpha plus beta, add up to pi  over two. Now, this formula is, you know,  interesting in its own right. But think about what it means  differentiably. I just calculated the derivative of arc  sine and I'm wondering, what's the derivative of arc cosine?  Well, whatever it is, right? These two functions add up to a constant, they're  pi over two. So, however wiggling y must affect arc  sine, wiggling y must affect arc cosine in precisely the opposite way so as to  make this sum a constant. Said differently, right?  The derivative of this side has to be the derivative of this side.  But the derivative of this side is zero, that means the derivative of arc sine  plus the derivative of arc cosine must be zero.  That means that the derivative of arc sine has to be exactly the negative of  the derivative of arc cosine. And since we know the derivative arc sine  is one over the square root of one minus x squared, that means the derivative of  arc cosine is negative one over the square root of one minus x squared.  My goal right now is to differentiate arc tan x, but that's too hard.  So instead, we're going to sort of sneak up on the derivative of arc tan x and  figure out the derivative without explicitly knowing it to begin with. So, here we go. I'll make this a little bit easier, I  don't want to keep writing arc tan x all the time.  So, let's give it a different name. let's just call it f.  So f of x will be arc tan x for the time being.  Now, what do I know about arc tan? Arc tan is one of these inverse  trigonometric functions. It's the inverse to tan.  So, I know what f of tangent x is. Well, as long as x is between minus pi  over four and pi over four, the arc tangent of tangent is just x.  Now, this is a really great thing, right? Now if, if f were differentiable, I could  use the chain rule on this. So, let's just pretend that arc tan is differentiable and see what happens. So, in that case, I differentiate this  using the chain rule. I got the derivative of f at the inside,  times the derivative of the inside. The derivative of tangent is secant squared. The derivative of x is one,  right? So, I differentiated both sides of this  equation, f tangent x equal to x.  The derivative of f tangent x is f prime tangent x times the derivative of tangent  secant squared. And the derivative of x, with respect to  x, is just one. Now, I'll divide both sides by secant squared x. And I find out that f prime of tangent x  is one over secant squared x. Now, what do I do? I, I'm trying to write down the derivative of f.  f is arc tan, and I want to know how to differentiate arc tan. and what I've  learned is that if f were differentiable, its derivative at tangent x would be one over secant squared x. Well, here's a trick.  I know the Pythagorean identity. I know that sine squared plus cosine  squared is one. So, if I take this identity and divide  both sides by cosine squared, I get that tangent squared,  right? That's sine squared over cosine squared, plus cosine squared over cosine  squared one, is one over cosine squared, that's a  secant squared. So, here's another trig identity, tangent  squared x plus one is secant squared x. So, up here where I've got one over  secant squared x, I could have written this as one over, instead of secant  squared x, I'll use the fact that secant squared x is tangent squared x plus one.  I'll put that in here and I get that the derivative of f, the derivative arc tan,  at tangent x is one over tangent squared x plus one.  Now, this is a pretty great formula, right?  This is saying that the derivative at something is one over that same thing  squared plus one. So a more normal, a more common way of  seeing this written is that the derivative of arc tan x is one over x  squared plus one. So, here's what we've done.  We've got the derivative of arc sine, is one over the square root of one minus x  squared. The derivative of arc cosine is exactly  negative that, because the sum of arc sine and arc cosine is a constant,  -1 over the squared of one minus x squared.  And we've calculated the derivative of arc tangent just now to be one over one  plus x squared. Excellent.  At this point, we've calculated the derivatives of arc sine, arc cosine, and  arc tangent. And now I challenge you to go and  calculate the derivatives of arc secant, arc cosecant, and arc cotangent.  

Why do sine and cosine oscillate?

>> . Sine is a periodic function. You follow the graph of sine just wiggles up and down, forever and ever. Let's try to get a sense as to why this is the case. Why does the graph of sine look like this? Why is sine moving up and down? To get some intuition for this, let's first imagine a very different situation. Let's suppose that velocity is equal to position. And, I gotta imagine I'm starting here, and I start moving, right? And I'm going to start moving slowly at first, but as my position increases, I'm going to be moving faster and faster, right? As my position is larger and larger, my velocity becomes larger and larger, which then makes my position larger and larger. So if this is the way the world works, I'm not bouncing around from minus 1 to 1 like sine. I'm just being thrown off towards infinity. We know what that situation looks like. If f of t is my position at time t, then f prime of t, the derivative of f at t is my velocity at time t. So saying that velocity equals position is just saying I've got some function whose derivative is equal to itself. And, we know an example of that. F of t equals e to the t is one example of such a function. So, to cook up a different situation, let's imagine something a little bit different. Instead, suppose that we're in the situation where my acceleration is negative my position. So that means that if I'm standing over here, where my position is positive, I'm to the right of zero, then my acceleration is negative and I'm being pulled back towards zero. And then, as I swing past toward zero, now my position is negative, so my acceleration is positive, so I'm being pulled back in the other direction. And that's kind of producing this swinging back and forth across 0. This situation comes up physically in the real world. If you attach a spring to the ceiling, the spring's acceleration is proportional to negative its displacement. There's a function also that realizes this model. To say that acceleration is negative position is as it is the second derivative of my position, which is my acceleration, is negative my function. And yeah, we know an example. If f is sine of t, then the derivative of sin is cosine. And the derivative of cosine, which is the second derivative of sine, is minus sine of t. And we know another function just like this. Look if, if g of t is cosine of t, then g prime of t is minus sine of t. But then, g double prime of t, the second derivative of cosine, is just minus cosine of t, because the derivative of sine is cosine. So cosine is another example of a function whose second derivative is negative itself. We know even more functions like this. For instance, what if f were the function f of t equals 17 sine t minus 17 cosine t? Well then, the derivative of f would be 17 times the derivative of sine, which is cosine, minus 17 times the derivative of cosine. But the derivative of cosine is minus sine. So this is plus 17 sine t. So that's the derivative of f. The second derivative of f is the derivative of the derivative. This'll be 17 times the derivative cosine. Which is minus sin, plus the derivative of sin, which is cosine. So yeah, this is another example of a function whose second derivative is negative itself. If I take this function and differentiate it twice, I get negative the original function. So if you like, the reason why all of these functions are bouncing up and down like this, is because in every case, the functions second derivative is negative its value. When the function is positive, the second derivative is negative, pulling it down. And when the function's value is negative, the second derivative is positive, pushing it back up. And so the function bounces up and down like this forever. 

How can we get a formula for sin(a+b)?

How do we really know that, that angle sum formula is true? What happens if I just rotate a single point? For instance, if I take this point here, the point (1, 0), and I rotate it through an angle of theta, then I end up at the point (cosine theta, sine theta). Somewhat similarly, if I take this point here, the point (0, 1), and I rotate counterclockwise through an angle of theta, I end up at the point (minus sine theta, cosine theta). What if I rotate some other point (x, 0)? Well, if I rotate (x, 0), say x is some number between 0 and 1, well, then I just think about where does (1, 0) go, right? (1,0) rotated at (cosine theta, sine theta). So (x, 0) must rotate to (x cosine theta, x sine theta), a point on the middle of this red line. What if I rotate (0, y) through an angle of theta? Similarly, if I've got some point, (0, y), say y is between 0 and 1, some point on this red line, if i rotate that point through an angle theta, well the point (0, 1) rotates over to the point (minus sine theta, cosine theta) on the circle. So the point (0, y) rotates to a point on the red line with coordinates (minus y sine theta, y cosine theta). All right? Its exactly this point, but scaled by y. Okay, so we know how to rotate (x, 0), and we know how to rotate (0,y). How do i rotate (x, y)? So, what if i rotate this point, (x, y), through an angle theta? Well, lets rotate this, and all i really have to figure out is, where does the point (x, 0) end up, and where does the point (0, y) end up? Well, (x, 0) ends up at (x cosine theta, x sine theta). This point started off as (x, 0), and when I rotate that through an angle theta, this is where it ends up. And this point started off as (0, y), and if I rotate that point through an angle of theta, this point ends up at (minus y sine theta, y cosine theta). Now where does this point, (x, y) end up? All I have to do is just add together these two coordinates to figure out the coordinates of this other point. And that means that this point, (x, y) ends up at x cosine theta minus y sine , and its y coordinate is x sine theta plus y cosine theta. So now i know how to rotate a point. How does that help? The goal was to try to justify the angle sum formula. The trick is to think about what happens when you do successive rotations. If you first rotate through and angle alpha, and then an angle beta, that's the same as rotating through an angle alpha plus beta all at once. Specifically, if I start with a point, (1, 0), and rotate through an angle of alpha, it ends up at (cosine alpha, sine alpha). I could then take that point and rotate it to the angle beta, and I've got a formula for that. Once I do that, the new coordinates are cosine alpha cosine beta minus sine alpha sine beta, and then the y-coordinate is cosine alpha sine beta plus sine alpha cosine beta. But that's the same point as I'd get if I started with (1, 0), and just rotated the whole thing initially through the angle alpha plus beta. This means that cosine alpha plus beta must be equal to cosine alpha cosine beta minus sine alpha sine beta, and sine alpha plus beta must be cosine alpha sine beta plus sine alpha cosine beta. These angle sum formulas can look very mysterious when they're written down, but they're really coming from a very simple source. They're coming from the fact that rotating through one angle and another angle is the same as rotating through the sum of those two angles. 

How can I approximate sin 1?

>> How can anyone compute sine? I mean nowadays, people use computers or caluculators. But by what mysterious force are these computational devices able to perform these computations? How does the machine know what sine of one is equal to? So, that's a good question. How do we compute sine of 1 radian? One way is to use a little bit of calculus. So, what does calculus tell us? So let's let f of equals sine x, and the derivative of x is cosine x, the value of f at 0 is 0, sine 00, and the value of the derivative at 0 is 1, because cosine 0 is 1. Now, I can use the derivative to approximate the functions value. The functions value near zero is the functions value at zero, plus how much I would be the input by, times the ratio of output change to input change, the derivative at zero. Well given this, in our specific case the value of function 0 and derivative at 0 is 1, and that means the functions value at h, for h very small, is about h. That means that sine of a very small angle is very close to that same value. Let's actually use some numbers. So very concretely, sine of say 1 32nd is really close to 1 32nd. Now, in addition to this, we've also got a double angle formula. The double angle formula for sine says that sine of 2x is 2 times sine x times cosine x. You can derive this formula from the angle sum formula for sine. Sine of x plus x gives you this. Alright. Another way to rewrite that a double angle formula is to write it entirely in terms of sine. Alright. Cosine is the square root of 1 minus sine squared x. Well, as long as the cosine is positive, this is true. So, if I'm working with a value of x, which cos sins positive, then sin of 2 x is 2 times sin x times the square root of 1 minus sin squared x. We can put these pieces together. Well, here's what we do. I know that sin of 1 32nd is very close to 1 32nd because 1 32nd is close to 0. And, sin of a number close to zero is about that. That same number. Then I can use this double-angle formula entirely in terms of sign to get an approximate value for sign of 1 16th by using my approximate value for sign of 1 32nd. Then, I can use this double-angle formula again to get an approximate value for sign of 1 18th, and again to get an approximate value for sign of 1 4th, and again to get an approximate value for sign of 1 half, and then again to get an approximate value of sign of 1. Which is really quite close to the actual value of sine of 1. So, what happened? Calculus told us that sine of a small number, is close to that small number. The double angle formula let us transport that bit of information around zero to points much further away. All the way to approximate sine of 1.