Derivative of Secx

Derivative of Secx

Sec(x) Derivative Rule

Secant is the reciprocal of the cosine. The secant of an angle designated by a variable x is notated as sec(x).
The derivative rule for sec(x) is given as:

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants sec x —1 and m and n are integers): Sec X + sec x —1 Let f (x) = sec x +1 Let u — — sec x So, f'(x) = Using quotient rule – sec x +1
Derivative of Secx
Derivative of Secx

let’s find the derivative…

of the secant of x.

How would we go about doing this?

Well sec(x) is another way of writing 1/cos(x)…

and since we have a quotient here,

the quotient rule would aptly apply…

so written in short hand, the quotient rule is:

y-dash or y-prime… I’m going to call it y-dash…
is equal to (vu’ – uv’) / v^2.

And in this case, we have y = sec(x), and…

I’m going to let u equal the numerator of 1

and v equal the denominator: cos(x).

So now if we apply the quotient rule…

for the first part, we have v, so cos(x) remains unchanged…

multiplied by the derivative of u.

So the derivative of 1, a constant, is zero…

then minus u, which is one, times the derivative of v…

which is the derivative of cos(x)…

and that’s equal to -sin(x)…

all over v^2, which is cos^2(x).

So the result is positive sin(x)…

divided by the cos^2(x).

And I can rewrite this as…

[sin(x)/cos(x)]*1/cos(x).

And then I can realise that this part is equal to…

the tangent of x… and this part is equal to again…

the secant of x!

So the derivative of sec(x) is equal to…

tan(x)*sec(x)

 

Derivative of sec(x)

Sec(x) Derivative Rule

Secant is the reciprocal of the cosine. The secant of an angle designated by a variable x is notated as sec(x).
The derivative rule for sec(x) is given as:

ddxsec(x) = tan(x)sec(x)

This derivative rule gives us the ability to quickly and directly differentiate sec(x).
X may be substituted for any other variable.
For example, the derivative ddysec(y) = tan(y)sec(y), and the derivative ddzsec(z) = tan(z)sec(z).

Proof of the Derivative Rule

The sec(x) derivative rule is originates from the relation that sec(x) = 1/cos(x). Now, the first step of finding the derivative of 1/cos(x) is using the quotient rule.

Using the quotient rule on 1/cos(x) gives us:
[sin(x)/cos(x)][1/cos(x)]
sin(x)/cos(x) = tan(x), and 1/cos(x) = sec(x)
Therefore, it simplifies to tan(x)sec(x), resulting in:
ddxsec(x) = tan(x)sec(x)

Derivative Rules of the other Trigonometry Functions

Derivative Rules Here’s the derivative rules for the other five major trig functions:
ddxsin(x) = cos(x)
ddxcos(x) = -sin(x)
ddxtan(x) = sec2(x)
ddxcot(x) = -csc2(x)
ddxcsc(x) = -cot(x)csc(x)

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Antiderivative of secxtanx
Antiderivative of secxtanx

We have to find the antiderivative of secx tanx. To find the antiderivative of secx
tanx we must first understand the meaning or definition of antiderivative.
Antiderivative is an operation which is opposite of derivation operation that
means antiderivative calculates the integral of a derivative.
Suppose we have a function f(x), its derivative is g(x) means d(f(x))= g(x) than
antiderivative of g(x) is f(x) that is J g(x) dx = f(x) + c dx. So we get f(x) as
antiderivative of g(x) is a constant.
We have understood the meaning of antiderivative, so now we will calculate the
antiderivative of secx tanx.
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The process of finding antiderivative of sec(x) tan(x) is shown in steps below-
Step 1 : Find the derivative of sec(x), that is d(sec x) / dx = sec(x) tan(x) dx.
Therefore g(x) sec(x) tan(x).
Step 2 : Now to find the antiderivative of sec(x) tan(x) , we will have to find the
integral of sec(x) tan(x).
That is Jsec(x) tan(x) dx = sec(x).
Proof of above operation- => Let g(x)= sec x tan x dx => g(x) = J sin x / cos 2 x
dx ( sec x = I / cos x , tan x = sin x / cos x) Let u = cosx
Therefore du – sin x dx g(x) – f 1 / u 2 du g(x) – u-2 du g(x) = 1/
u + C => g(x) = 1/ cos x + C => g(x) sec x + C Hence proved. Here we have
used method of integration by parts to calculate the antiderivative of sec x tan x.
Hence to find the antiderivative of secxtanx we have to calculate the integral of

Derivative of cosx

Derivative of cosx

Derivative of cosx  We see that number 2 that FX is equal to Cos X the next step is F of X plus h is equal to cos of X plus h then by first principle of derivative we have F dash X is equal to repeat H tends to 0 f of X plus h minus FX totally divided by H not equal to limit H tends to 0 for f of X plus h we write cos of X plus h minus 4 F X we substitute kha’zix and all thing divided by H.

Derivative of cosx sin x and cos x
Derivative of cosx sin x and cos x

now here we apply the formula cos C minus cos D so that equal to minus to sign in brackets C plus D upon 2 into again sign in bracket C minus D upon to say in present case X plus h represents C and X represent D so the next step is is equal to the limit H tends to 0 minus to sign in bracket X plus h plus X upon 2 into sign in bracket X plus h minus X upon 2 and then whole thing divided by H so the next step will be limit H tends to 0 minus 2 into sign in bracket to express H upon 2 into the limit X tends to 0 sign in bracket we cancel plus X minus 6 so we left me H by 2 and in the denominator we have H.

Derivative of cosx

so that we write H by 2 to get sine rule and into 1 by 2 so this is require adjustment so that equal to minus 2 we write as it is into sine in bracket 2 x 4 h we substitute 0 as a limit upon 2 into now let me extend to 0 sine of H by 2 upon H by 2 that gives 1 into 1 by 2 now you cancel – – from dominator denominator and again to find 2 from numerator denominator inside sine function so we left with only – sine X so in this way we prove a brush X is equal to minus sine X it means the derivative of cos x is minus sine X

Zero product property

Derivative of cosx sin x and cos x

limit definition of derivative

Derivative of cosx sin x and cos x

Derivative of cosx sin x and cos x

Derivative of cosx sin x and cos x

The trigonometric functions \sin(x)sin(x)sine, left parenthesis, x, right parenthesis and \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis play a significant role in calculus. These are their derivatives:
\begin{aligned} \dfrac{d}{dx}[\sin(x)]&=\cos(x) \\\\ \dfrac{d}{dx}[\cos(x)]&=-\sin(x) \end{aligned}
dx
d

[sin(x)]
dx
d

[cos(x)]

=cos(x)
=−sin(x)

Derivative of cosx sin x and cos x
The AP Calculus course doesn’t require knowing the proofs of these derivatives, but we believe that as long as a proof is accessible, there’s always something to (Derivative of cosx sin x and cos x ) learn from it. In general, it’s always good to require some kind of proof or justification for the theorems you learn.
First, we would like to find two tricky limits that are used in our proof.
1. \displaystyle\lim_{x\to 0}\dfrac{\sin(x)}{x}=1
x→0
lim

x
sin(x)

=1limit, start subscript, x, \to, 0, end subscript, start fraction, sine, left parenthesis, x, right parenthesis, divided by, x, end fraction, equals, 1
Khan Academy video wrapper
2. \displaystyle\lim_{x\to 0}\dfrac{1-\cos(x)}{x}=0
x→0
lim

x
1−cos(x)

=0limit, start subscript, x, \to, 0, end subscript, start fraction, 1, minus, cosine, left parenthesis, x, right parenthesis, divided by, x, end fraction, equals, 0
Khan Academy video wrapper
Now we are ready to prove that the derivative of \sin(x)sin(x)sine, left parenthesis, x, right parenthesis is \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis.
Khan Academy video wrapper
Finally, we can use the fact that the derivative of \sin(x)sin(x)sine, left parenthesis, x, right parenthesis is \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis to show that the derivative of \cos(x)cos(x)cosine, left parenthesis, x, right parenthesis is -\sin(x)−sin(x)minus, sine, left parenthesis, x, right parenthesis.

Derivative of cosx sin x and cos x, the derivatives, of two of thegreat functions of mathematics: sinex and cosine x. Why do I say great functions? What sort of motion do wesee sines and cosines? Well, I guess I’m thinkingof oscillations. Things go back and forth. They go up and down. They go round in a circle.

Your heart beats andbeats and beats. Your lungs go in and out. The earth goes around the sun. So many motions are repeatingmotions, and that’s when sines and cosines show up. The opposite is growingmotions. That’s where we have powers ofx, x cubed, x to the n-th. Or if we really wantthe motion to get going, e to the x. Or decaying would bee to the minus x. So there are two kinds here.

 

Derivative of cosx sin x and cos x
Derivative of cosx sin x and cos x

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Derivative of cosx sin x and cos x

We’re talking about the onesthat repeat and stay level, and they all involvesines and cosines. And to make that point,I’m going to have to– you know what sines and cosinesare for triangles from trigonometry. But I have to make thosetriangles move. So I’m going to put the trianglein (Derivative of cosx sin x and cos x )  a circle, with one corner at the center, andanother corner on the circle, and I’m going to movethat point. So it’s going to becircular motion. It’s going to be themotion that– the perfect model of repeatingmotion, around and around the circle. And then the answer we’re goingto get is just great. The derivative of sine x turnsout to be cosine x.

And the derivative of cosine xturns out to be minus sine x. You couldn’t ask for more. So my interest is always toexplain those, but then I want to really– we’re seeing this limit stuffin taking a derivative, and here’s a chance for meto find a limit. This turns out to be the crucialquantity: the sine of an angle divided by the angle,when the angle goes to 0. Of course, when it’s at 0,the sine of 0 is 0, so we have 0 over 0. This is the big problemof calculus. You can’t be at the limit,because it’s 0 over 0 at that point. But you can be close to it. And then if we drew a graph, hada calculator, whatever we do, we would see that that ratiois very close to 1, but today we’re going to actuallyprove it from the meaning of sine theta.

Now remember whatthat meaning is. So back to the startof the world. Actually back to Pythagoras,way, way back. The key fact is what youremember about right triangles, a squared plus bsquared equals c squared. That’s where everything startsfor a right triangle. I don’t know (Derivative of cosx sin x and cos x ) if Pythagorasknew how to prove it. I think his friendshelped him. A lot of people havesuggested proofs. Einstein gave a proof. Some US president evengave a proof. So it’s a fundamental fact, andI’m going to divide by c squared, because I’d like theright hand side to be 1. So if I divide by c squared, Ijust have a squared over c squared plus b squaredover c squared is 1. And I’m going to make thathypotenuse in my picture 1.

So then this will be the a overc, and that ratio of the near side to the hypotenuseis the cosine. So what I have here iscosine theta squared. Let me put theta in there. Theta is that angleat the center. And what’s b? So this is a over c. That’s cosine theta. B over c is this point,and that’s sine theta. And they add to 1. So that’s Pythagoras usingsines and cosines. So this is the cosine. And this vertical distanceis sine theta. OK, so that’s thetriangle I like. That’s the triangle that’sgoing to move. As this point goes around thecircle at a steady speed, this triangle is going to move. The base will go left andright, left and right.

The height will go up and down,up and down, following cosine and sine. And we want to know thingsabout the speed. OK, so that’s circular motion. Now I’ve introducedthis word radians. And let me remind you what theyare and why we need them. Why don’t we just (Derivative of cosx sin x and cos x ) use 360degrees for the full circle? 360 degrees. Well, that’s a nicenumber, 360. Somebody must have thought itwas really nice, and chose it for measuring anglesaround the world. It’s nice, but it’snot natural.

Somebody thought of it,so it’s not good. What we need is the naturalway to measure the angle. If we don’t use the natural way,then this is the sine– if I measure this xin degrees, that formula won’t be right. There will be a miserable factorthat I want to be 1. So I have to measure the anglesthe right way, and here’s the idea of radians. The measure of thatangle is this distance around the circle. That distance I’m going to calltheta, and I’m going to say this angle is thetaradians when that distance is theta. So that now, what’sa full circle? A full circle would mean theangle went all the way round. I get the whole (Derivative of cosx sin x and cos x )  circumference,which is 2 pi. So 360 degrees is2 pi radians. So the natural numberhere is 2 pi. This can’t be helped, it’sthe right one to use. Radians are the right wayto measure an angle. So now I’m ready to do the jobof finding this derivative.

OK. Let me start at thekey point 0. If we get this one, we getall the rest easily. So I’m looking at the graphof the sine curve. I’m starting at 0. We know what sine theta lookslike, and I’m interested in the slope, the derivative. That’s what this subjectis about, calculus, differentiating. So I want to know theslope at that point. And it’s 1. And how do we showthat it’s 1? So now I’m coming to the pointwhere I’m going to give a proof that is 1. And the proof isn’t just for thesake of formality or rigor or something. You really have to understandthe sine function, the cosine function, and this isthe heart of it. OK.

So we want to showthat slope is 1. How am I going to do that? That’s the slope, right? If I go a tiny amount theta,then I go up sine theta. So in this average slope, ifI take a finite step– I could have called it deltatheta, but I don’t want to write deltas all the time. So I just go out a littledistance theta and up to the sine curve. I stopped at the sinecurve by the way. The straight line is a littleabove the sine curve here. And that ratio, up divided byacross, that’s the delta sine divided by delta theta. And because it started at 0,it’s just sine theta is the distance up, and theta isthe distance across. So this is the average slope.

And of course you rememberwhat calculus is doing. There’s always this limitingprocess where you push things closer and closer to the point,and you find the (Derivative of cosx sin x and cos x ) slope at that point, sometimes calledthe instantaneous velocity or slopeor derivative. Now here’s the way it’sgoing to work. I’m going to show that sinetheta over theta is always below 1. So two facts I want to prove. I want to show that sine thetaover theta is less– sorry, sine theta over theta– well, let me get this right. I might as well putit the neat way. I want to show that sinetheta is below theta. This is for thetagreater than 0. That’s what I’m doing. OK. So that tells me that thiscurve stays under that straight line, that 45 degreeline, which I’m claiming is the tangent line.

 

And it tells me when I divideit by the theta, it tells me that sine theta overtheta is below 1. But now how much below 1? Right now if I only know this,I haven’t ruled out the possibility that the slopecould be much smaller. So I need something below it. And fantastically, thecosine is below it. So the other thing that I wantto prove is that the cosine– and I’ll let me doit this way. I’m going to show the tangent oftheta is bigger than theta. Again, some range of thetas. Positive thetas upto somewhere. I don’t know, (Derivative of cosx sin x and cos x ) I thinkmaybe pi over 2. But the main point is near0, that’s the main point. So can I just rewrite– do you remember whatthe tangent is? Of course, sine thetaover cosine theta. So this is sine thetaover cosine theta, bigger than theta. We still have to prove this. And now I want to bring thetheta down and move the cosine up, and that will tell me thatsine theta, when I divide by the theta and multiplyby the cosine theta. So that was the same as that,was the same as that. And that’s what I want. That tells that this ratio isabove the cosine curve.

 

Do you see that if I canconvince you, and convince myself that these are both true,that this picture is right, then– I haven’t gone into gorydetail about limits. If you really insist,I’ll do it later. But whatever. You can see this has just got tobe true, that if this curve is squeezed between the cosinecurve and the 1, then as theta gets smaller and smaller,it’s squeezed to equal 1 in the limit. Allow (Derivative of cosx sin x and cos x ) me to say that that’spretty darn clear. OK. Whatever limit meets. So these are the main factsthat I need to show. And I need to show thoseusing trig, right? I have to draw some graphthat convinces you. And this isn’t quite goodenough, because I just sketched a sine graph. I have to say where doessine theta come from? OK.

 

So this will be number one, andthis will be number two, and when we get those two thingsconvincing, then we know that sine theta over thetais squeezed between and approaches 1. And then we’ll know the storyat the start, and you’ll see that it becomes easy tofind these formulas all along the curve. OK. Ready for these two? Number one and number two. OK, number one. Why is sine theta– oh, I can probably seeit on this picture. Yeah, I can prove numberone on this picture. Look, that piece wassine theta, right? And I want to provethat this length– what am I trying to prove? That sine theta isbelow theta. Let me write it againwhat it is to show. In math it’s always a goodidea to keep reminding yourself of what itis you’re doing. Sine theta is below theta. OK. So why is it? And you see it here. That was sine theta, right? And where was theta? Well, because we measured thetain radians, theta is this curvy distance that’sclearly longer.

 

The shortest way from this tothe axis there is straight down, and that’s sine theta. A slower way is go round andend up at not the nearest point, and that was theta. Is that good? I could sometimes just to beeven more convincing, you add a second angle, and you say OK,there’s 2 sine thetas and here is 2 thetas, and clearly weall know that the shortest way from there to thereis the straight way. So I regard this as doneby that picture. You see we didn’tjust make it up. It went back to the fundamentalidea of where sine theta is in a picture. Now I need another picture. Yeah, I need another picturefor number two to show that tan theta is biggerthan theta. That was our other job. So essentially, I need thatsame picture again. Whoops, let me drawthat triangle. Yeah, and it’s got a circle. OK, that’s not a bad circle. It’s got an angle theta. And now I’m going to– math has always gotsome little trick. So this is it. Go all the way out, so now thebase is 1, and this is still the angle theta. And what else do I knowon that picture? Now I’ve scaled the triangleup from this little one, so the base is 1.

So what’s that height? Well, the ratio of the oppositeside to the near side, that’s what tangent is. Tangent is the ratio– whatever size the triangle– is the ratio of the oppositeside to the near side. Sine to cosine, here it’stan theta is that distance, and to 1. Good. OK, but now how am Igoing to see this? I have to ask you–and it’s OK– to think about area insteadof distance for a moment. What about area? So what do I see of area? I see right away that the areaof this triangle is smaller than the area– sorry, I shouldn’t have calledthat a triangle. That’s a little piece of pie,a little sector of a circle. So the area of this shaded– did I shade it OK– is less than– so this is the areaof the sector. Can I just call it the pie,piece of a pie, is less than the area of the triangle. But we know what the areaof the triangle is. What’s the area of a triangle? We can do that. It’s the base half, right? 1/2 times the basetimes the height. So the area of the triangleis 1/2 times the base 1 times the height. OK. Notice we’ve got the signgoing the right way. We want tan theta to bebigger than something,

 

so what do I hope? I hope that the area of thisshaded part, the area of the circular sector, is1/2 of theta. Wouldn’t that be wonderful? If I look at those areas,nobody’s in any doubt that this piece, this sector that’sinside the triangle, has an area less than the areaof the triangle. So now I just have to rememberwhy is the area of this sector, half of theta. You know, there’s another reasonwhy areas come up right when we use a radians, when wemeasure theta with radians. So remember, just think aboutthis piece of pie compared to the whole pie. What’s the area of thewhole piece of pie? So I’m explaining 1/2 theta. The area of the whole pie– I’m going to get some terriblepun here on the word pie. Unintended, forgive it. The area of that whole circle,the radius is 1, we all know what the area of a circleis pi r squared. r is 1, so the area is pi. My God, I didn’t expect that. Now what about this? What fraction is this sector? Well, the whole angle would be2 pi, and this part of it is theta, so I have the sector istheta over 2 pi, that’s the angle fraction, times thepi, the whole area. Do you see it? This piece of pie, orpizza, whatever– yeah, if I’d said pizza, Iwouldn’t have had that terrible pun.

Forget it. So the area of this piece ofpizza compared to the whole one is theta overthe whole 2 pi. Suppose it was a pizza cutin the usual 6 pieces. Then this would be a 60degree angle, but I don’t want degrees. What would be the angle of thatpiece of pizza that’s cut when the whole pizza’scut in 6? It would be 1/6 of 360. That’s 60 degrees. But I don’t want degrees. It’s 1/6 of 2 pi. And this one is theta of 2 pi. Anyway, the pis cancel. Theta over 2 is the rightanswer, and now we can cancel the 1/2, and we’vegot what we want. That’s pretty nice when yourealize that we were facing for the first time, more orless, the sort of tough problem of calculus when I can’treally divide theta into sine theta. Sine theta, I can’tjust divide it in. I have to keep them bothapproaching 0 over 0, and see what that ratio is doing. And now I said to conclude,I’ll go back and prove the slopes, find the slopesat all points. OK, so at all points– now let’s start with sine x.

So what am I doing now? I’m looking at the sine curve. You remember it went up likethis and down like this. I’m taking any point x. Suddenly I changed the anglefrom theta to x, just because I’m used to functions of x. We’re just talkingletters there. X is good, and this isa graph of sine x. X is measured in radians still. OK. So now what am I doing to findthe derivative at some particular point? I look at the sine there. I go a little distancedelta x. I go up to here, andI look to see– I want to know the changein sine x divided by the change in x. And of course, I’m goingto let the piece get smaller and smaller. That’s what calculus does. The main point is my x isnow here instead of being at the start. I’ve done it for the start, butnow I have to do it for all the other x’s. So there’s the x. There’s the x plus deltax, a little bit long. In other words, can I write thisin the familiar way, sine of x plus delta minus sinethere divided by delta x? OK.

 

So again, we can’t simplifytotally by just dividing the delta x in. We’ve got to go backto trigonometry. Trig had a formula forthe sine of a plus b. Two angles added, then there’sa neat formula for it. So the sine– can I remind you ofthat formula? It is the sine of the firstangle times the cosine of the second minus the cosine of thefirst angle times the sine of the second. OK? You remember this,right, from trig? The sine of a plus bis this neat thing. Now I have to subtract sine x. So now can I subtractoff sine x? When I subtract off sine x,then I need a minus 1. And now I have to divideby delta x. So I divide this by delta x, andI divide this by delta x. OK. This is an expressionI can work with. That’s why I had to rememberthis trig formula to get this expression that Ican work with. Why do I say I canwork with it? Because this is exactly whatI’ve already pinned down. Delta x is now headed for 0. This point is going to comeclose to this one.

So actually, I’ve got two terms.This sine delta x over delta x, what does that doas delta x goes to 0? It goes to 1. That was the point of thatwhole right hand board. So this thing goes to 1. Wait a minute. That’s a plus sign. Everybody watching is goingto think, OK, forgot trig. The sine of the sum of an angleis the sine times the cosine plus the cosinetimes the sine. Sorry about that one too. OK, so sine of delta x overdelta x goes to 0. And now finally, this goes to 1,and actually I need another little piece. I need to know this piece, andI need to know that that ratio goes to 0. So I need to go back to thatboard and look again at the cosine curve at 0. Because this will be a slopeof the cosine curve at 0. And the slope comes out 0for the cosine curve. The slope for the sinecurve came out 1.

Do you see how it’s working? So this is gone becauseof the 0. This is the cosine x times 1. All together I getcosine of x. Hooray. That’s the goal. That’s the predicted plan,desired formula cos x for the ratio of delta of sine x overdelta x as delta x goes to 0. Do you see that? So we used a trig formula,and we got the sine right a little late. Well, of course the reason I– one reason I goofed was that theother example, the other case we need for thesecond formula does have a minus sign. And it survives in the end. So I would do exactly the samething for the cosines that I did for the sines. If there’s another boardunderneath here, I’m going to do it. Yeah, there is. Now I want to know the deltaof cosine x over delta x.

That’s what we do, we have tosimplify that, then we have to let delta x go to 0. So what does this mean? This means the cosine a littlebit along minus the cosine at the point divided by delta x. Again, we can’t do that divisionjust right away, but we can simplify thisby remembering the formula that cosign– now let me try to remember it. It’s a cosine times a cosinefor this guy plus a sine– no, minus a sinetimes the sine. That’s the formula that we allremember and go to sleep with. Now divide by delta x. Oh, first subtract cosine x. So there was a cosine x, so Iwant to subtract one of them. OK?

And now I have to divideby the delta x. So I do that there. I do it here. And you see that we’re inthe same happy position. We’re in the happy position thatas delta x goes to 0, we know what this does. That goes to 1. We know what this does,or we soon will. That goes to 0, just theway they did on the board that got raised. So that term disappearedjust like before. This term survives. It’s got a 1, (Derivative of cosx sin x and cos x )  it’s got now asine x, and it’s got now a minus sign. So that’s the final result,that the limit is minus sine x. That’s the slope ofthe cosine curve. And you wouldn’t wantit any other way. You want that minus sign. You’ll see it with secondderivatives. So it’s just terrific thatthose functions, the derivative of the sine is thecosine with a plus, the derivative of the cosine isthe sine with a minus.

 

OK. And we’ve almost proved it, wejust didn’t quite pick up this point yet, and let me do that. That will finish this lecture. Why does that ratioapproach zero? What is that ratio? That ratio is coming fromthe cosine curve. The cosine curve at 0, the waythis ratio came from the sine curve at 0. Here I’m taking– this is delta cosine. There’s lots of ways I can dothis, but maybe I’ll just do it the way you see it. What’s the slope ofthe cosine at 0? Yeah, I think I can ask thatwithout doing limits, without doing hard work. I’ll just add the rest of thecosine curve, because we know it’s symmetric. What’s the slopeat that point? This is actually the mostimportant application of calculus, is to locatethe place where a function has a maximum. The cosine, its maximumis right there. Its maximum value is 1, and ithappens at theta equals 0. So the slope at a maximum– all right, I’m goingto put this– I could get this result by thesepictures, but let me do it short circuit. The slope at the maximum is 0. OK. Your intuition tells you that. If the slope was positive,the function would still be rising. It wouldn’t be a maximum, itwould be going higher. If the slope was negative, thefunction would be coming down, and the maximum wouldhave been earlier.

 

But here the maximumis right there. The slope has to be0 at that point. And that’s the quantity that wewere after, because this is the cosine of delta x. There is the cosineof delta x. Here is the 1, here is the deltax, and this ratio is height over slope. It gets to height over slope aswe get closer and closer. It’s the derivative, andit’s 0 at a maximum. And my notes give another wayto convince yourself that that’s 0 (Derivative of cosx sin x and cos x ) by using these factsthat we’ve already got. OK. End of the– so let me justrecap one moment, which this board will do. We now know the derivativesof two of the great functions of calculus. We already know the derivativeof x to the n-th, and in the future is coming e to thex and the logarithm. Then you’ve got the big ones.

Zero product property

limit definition of derivative

Derivative of cosx sin x and cos x