Derivative of sin(x) and cos(x), PROOF
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- เผยแพร่เมื่อ 23 ส.ค. 2024
- Geometric proof of sin(x)/x approaches 1 as x approaches 0, • The Limit (do not use ...
Angle sum formula: • Angle sum identities f...
part1: derivative of sin(x) and cos(x), • Derivative of sin(x) a...
part2: derivative of tan(x) and cot(x), • derivatives of tan(x) ...
part3: derivative of sec(x) and csc(x), • derivative of sec(x) a...
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"sine and cosine are like homies" #yay
Yup!
Thats hawitt
I saw a student using L'Hospital for this, it made me really angry because he used sin'(x) to calculate sin'(x)
PackSciences Hypothetically it might be possible to use LH’s rule to simultaneously calculate sin’ and cos’ if it results in two expressions with two unknowns, much like using implicit differentiation where you have a derivative on both sides of an equation and solve for the derivative.
In this case you end up with sin’(x) = sin(x)cos’(0) + cos(x)sin’(0) and since cos(x+h) - cos (x) = cos(x)sin(h) - sin(x)cos(h) - cos(x) you get cos’(x) = cos(x)sin’(0) - sin(x)cos’(0). Whether or not you could use those two equations to simplify out the sin’(x) and cos’(x) values in terms of sin(x) and cos(x) is another question, but on the face of it this sort of method isn’t completely out of line in general.
It wouldn't be perfect wrong.
Not good to get angry at students for trying. Simply guide them in the right direction and explain why you can't use circular reasoning (to some, it isn't obvious and it should be explicitly stated nonetheless). Anger makes most people not want to learn fron you in a teaching setting.
leave your teaching carrier 👍🙂
I know this comment is 5 years old but this method asumes that the derivative exists in the first place, which might not be true.@@Bodyknock
Prefect! I didn't know that cosine stands for complement of sine. Thanks for the video!
Sine and Cosine are basically opposites.
It explains why the Tangent is the same as Sine/Cosine.
It also explains why the Tangent of 90 degrees is undefined as Sine of 90 is 1 and Cosine of 90 is 0, so 1/0 is undefined.
Student: I'm so smart, I know how to derive all the trig function derivatives
Bprp: Really? Can you show me it for sin and cos then
Student: ...
#yay
What's up with that hashtag
Can u do that proof of cosh-1/h and sinh/h
Sinx/x value is 1 and cosh-1/h substitute h value
Since you might know sinh/h is equal to 1 but for (cosh-1)/h we can solve it like
lim [h tends to 0] (cosh -1)/h
We can use the trigonometric function of cos2x just substitute 2x by h and we can break it into sine functions as cosh=1-2sin²(h/2)
So next we just substitute cosh in the above equation as
[1-2sin²(h/2)-1]/h
= -2sin²(h/2)/h
And now using limits
lim [h tends to 0] -2 [sin²(h/2)/(h/2)² × h/4
Again [sin(h/2)/(h/2)]² is equal to 1
Therefore lim [h tends to 0] -h/2 which after putting the value of the limit we get 0
You can also use the identity sin^2 + cos^2 = 1 and derive both sides then you got (sin^2 + cos^2) ‘ = 0 and (cos^2) ‘ = -(sin^2)’ so 2cos*(cos)’ = -2sincos and so (cos)’ = -sin
GODDAMMIT, IT'S THAT EASY?
May sound complicated but this is to show that you can use and play with identities to prove common relation in maths :)
@@elbonais683you take d/dx(sinx) = cosx for granted, but it's still cool
I ended with lim as h->0 cos(x)*sin(h/2)/(h/2). Using some tricks with the e^iz formula xD.
In summary, It's like 19÷4 = 19/4
Nice! I hadn't thought about using complementary identity to prove the derivative of cos(x)
Since we're all adults now... Tee-hee! 😁
?
South park
I am not an adult I'm 12
mario mario you are a math adult!
Master Cao, no explanation of why (cos(h)-1)/h tend to 0 when h tend to 0
Gaga Daddy 5:35 He is aware of this, but it would have been too much for this video to explain this in detail. I'm sure he is able to show this extra proof if he wants to.
@novidsonmychannel, hi! thank for your advise! I am not challenge Master Cao for ignoring the Limit part. The Point here is: lim(h->0) cos(h)-1/h goes to zero can be applied L'Hospital rule, easily. However, if we do so, it go back to the origin point - we want to work out the derivative of sin and cos function from fundamental. This is MY dilemma! ... and this' why I asked this question... sorry if there is any clever way which I am not aware!
Gaga Daddy No problem. ;) I can understand what you mean. And I admit that I don't know another "clever" way either. I only can sketch an idea of a "proof" for the two limits: We know that sin(0) = 0 and sin(h) is approximately equal to h for abs(h) 0 ((cos(h))'/1) = 0/1 = 0. I am aware that every mathematician would scream seeing this "proof", but since I'm studying a physical subject please forgive me. :P At least for me it is sufficient if I find ways like these to understand the mathematical backgrounds.
@novidsonmychannel, thank Physicist! Hope u be another Hall of Fame in your professional area! :)
You're welcome, thank you too! :)
Thanks a lot sir .
Amazing explaintion 😀
God bless you instead lecture was superb 👏🏻👏🏻
This is really neat! Glad I found this! =D
2:05 for those ho want to understand how he get the rule; go watch videos about addition and soustraction for cosinus and sinus cos(a+b) cos(a-b) sinus (a+b) sinus (a-b) it s kinda difficult but you will understand it ; then after that get back to the video
ty
Why not do the derivative based on the Maclaurin series for since and cosine? The approach you used here has the issue that, without the numerical ✋ waving you might have been stuck with L'Hospital's rule, essentially needing to know the answer to the derivative you were trying to find.
Maybe there's a non-circular, rigorous way to solve the "0/0" limits without L'Hospital's rule, but it didn't come readily to mind.
Sir, with due respect, people like me who are new to calculus and just learning the derivatives of the trig functions often wonder how these derivatives came about. And while this might not be the most rigorous proof out there, it is more accessible, and, as you say, may be proved to be rigorous. Which is why I'm grateful to BPRP for this video.
Please correct me if this is not the case: Don't the Maclaurin series for sin and cos require the result in the above proof to start with?
Can you prove
tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)]? BTW thanks so much, I learned much in your videos.
That directly follows from tan(x+y)= sin(x+y)/ cos(x+y)
Just give it a try.
@@leadnitrate2194 what about sin(x+y)??? its proof
@@itookashower3485 the proof requires a few illustrations, so I can't outline it in the comments.
But this video by bprp shows it very well
th-cam.com/video/2SlvKnlVx7U/w-d-xo.html
Hope it helps
@@itookashower3485 you can also prove it by writing sin θ= {e^(iθ) - e^(-iθ)}/2 and cos θ = {e^(iθ) + e^(-iθ)}/2 but I don't know if you've studied complex numbers yet.
Finally, a video that I can understand xD
Yes
#deathSTRoKEgamingAKANKSHYA
saving lives in 2022 T-T thank you for this
Best teacher in world ❤
I figured out two proofs that don’t use the limit of sin x/x, the limit of (cos x-1)/x, or any angle-sum identity. One uses the definition of arc length (as well as the Pythagorean theorem, the fundamental theorem of calculus, and the derivative of sqrt(1-x^2)), but the other one just uses the parametric definition of a derivative (d[x,y]/dt=[dx/dt,dy/dt]). If I ever teach a math class, I will be looking for one of those.
This is pretty neat. #yay
A more elegant and compact proof exists which uses the identity SinC - SinD = 2Sin(C-D)/2*Cos(C+D)/2 together with the limit Sinh/h ➡️ 1 as h ➡️ 0.
Note, C = (x + h) and D = x.
Substitution: lim h ➡️ 0 (2Sinh/2h)*(Cos((2x+h)/2)) evaluates to (1)*(Cos(2x/2)) which, in turn, evaluates to Cosx.
I understand good 😊😊😋😋
this video really reminded me on an older one of yours. you even said they're like homies. :)
#YAY
Really good
I wonder where trig identities come from, would you please explain us? Thanks :) #yay
Wierzbi sure. It's here th-cam.com/video/2SlvKnlVx7U/w-d-xo.html
Nice video!!
MathForLife thanks! And glad to see you back!!
blackpenredpen thanks! I was moving to Berkeley:)
MathForLife nice!!! How you like it there so far??
blackpenredpen I love it!! Everything is so close:)
the doraemon theme playing at the start is just awesome
You can also use the expansion of sinx and it is very easy with that approch
Hey brother could you please provide a geometrical proof ?(actually, I was anticipating for one such proof........ )
This is a very good video. I have the geometrical proof in my video here. th-cam.com/video/64dguvQBwUQ/w-d-xo.html
It is nice
Thank you so much
can you use the definition of the derivative for e^x??
e^x=1+x/1!+x^2/2!+x^3/3!+x^4/4!...
e^x=sum x^n/n! ]0-inf
Take derivative of that
You are my Best 👌 👍 😍
well for the definition of the derivative i tend to do a "useless" extra step. i write the limit but the denominator (anyone with eyes will see why it´s "useless") i´ll write down "x-x+h" ofc it is just h. i´m aware of that. but i just want that pair of f(x) in the numerator amd x in the denominator and f(x+h) in the denominator ans x+h in the numerator just to show that this is nothing but the slope of a line between two points. and then, when this is clear we can kill the x-x in the denominator like a sensible human being and get to work but okay i´m a maths tutor so i show it to kids. Not adults who have their own fair share of mathematical knowledge. sow there is no need to link it to previously learned things this strongly. YOur viewers would appreciate something brand new as well as something build on a foundation they already have.
Another alternative way to compute (cos(x))' if you already know (sin(x))', would also be to use cos(x) = sqrt(1-(sin(x))^2) and then the chain rule - works as well, I tried it. ;)
And of course you could also use the definition sin(x) = (1/(2i))(e^(-ix) - e^(ix)) by using the chain rule - if you are aware of the derivative of e^x and the definition cos(x) = (1/2)(e^(-ix)+e^(ix)) so that you can recognize it in your result. This is much faster, but of course you're aware of this and on the other hand I also like it to find non-complex proofs for real functions - especially if you explain it to students who don't already know complex numbers.
how to proof the lim for (cost(h)-1)/h is 0? 0/0=infinite
Sir
What is the answer of
-d/dx cos x
Pls reply me sir🙏🙏
Just a little question:
Couldn't you just say
cos(h) approaches h
and
sin(h) approaches (1+h)
?
The result is the right one but we got the answer in less steps. #yay
Could somebody please give me an answer? :o
Thank u sir ❤🎉
The limit of (f(x+h) - f(x)) /h
Where did it come from??
general definition
I use the complex definitions
Doremon theme song in the background , so gooooooooood
Thank sir for guiding
Why write cos(h) please tell me sir 1:59 video please explain Sir
Did you get it already? If not, I think it is because sin(x+h) has an equivalent identity which is sin(x)cos(h)+cos(x)sin(h)
In trigo, it is written as
sin(a±b) = sin(a)cos(b) ± cos(a)sin(b)
correct me if I'm wrong. thanks
You are undoubtedly a cool human! 😎
Thanks sir, well explained ❤️❤️❤️
What is your language sir but teaching mathod is very nice
Which company 's markers do u use??
Sir,
What is dα/ dx of sec α
Why can you bring the sin of x and the cosine of x out?
can you find the continuity (or not) of y=x^(1/x) from -inf to 0 ?
I love your videos about complex #
non-continuous, plug in x=-2
I don't think it's continuous because some are complex but at -1 it's not
@Jordan Saenz: y at -1 is also complex, so is at 1, 2, R & C
Dawid Krainski oh yeah it is oops
I still wonder how derivative of sinx can be cosx . Is it possible to proof the derivative of sinx is cosx from graph of it
Can you do an indefinite integral of cos(tan(sec(x)))? All calculators stuck on this.
no elementary function in terms of standard mathematical
Heinz Anderson what?
you can not present the solution in a closed form
Vilém Jankovský There are already more than enough more "simple" functions which have no explicit indefinite integral, in other words no anti-derivative in terms of elementary functions. e^(x^2), sin(x^2), 1/(ln(x)) - just to name some. Every calculator would stuck on these as well.
novidsonmychannel justcommenting Oh, thank you.
Amazing
I came here to understand a mug up 1step. But here he says to mug 10 steps ahh shit😂
how to prove derivative of f(x) = (u(x))^n?
sin and cos are like homies : )
Thank you ☺️
Sir can you solve d/dx(e^x sinx) ?
We can also do this using Series expansion of Sinx then taking derivative of Intial terms
Why was Doraemon theme playing on the background
I wonder, what the full derivative of sin(a+b) is, since the full derivative of a multivariable function is more than just the successive partial derivatives? Mainly since layering the partials would simply give -sin(a+b)
👍🏻👍🏻
i have a problem
can you solve for me?
solve for x
x+[x]=1
RUPAK BISWAS I think x€IR_
if x=0.5
[x]=0
than x+[x]=0.5
You can solve it by graph; it's no solution.
Thank
x+|x|=1 x+x=1 if x>0 and x-x=1 if x0 and 0=1 if x
Where are my homies?!!!
Pls someone explain me 3:43 how (sin(h)-1)/h become zero because when I calculated it on calculator the value show very large.
Very late answer, but its (cos(h) - 1)/h that approach 0, not (sin(h) - 1)/h
sinhcosx on the third line dont get it wrong guys
Homies......LOL
you saved me 5 marks sin my alevel thanks
Dude, but you did not proove the limits :(
It’s in another video. Search “the limit”
Deja-vu: th-cam.com/video/VMNX2xGffzU/w-d-xo.html
First!
Oscar Troncoso yay!!!
baby help me
:v