Proof: Derivative of Sin is Cos (Version 2)

Proof. visualization, and discussion on how the derivative of sin is cosine.

Created by David Longstreet, Professor of the Universe, MyBookSucks.Com

[physicsgod]

For true mathematician or physicist it’s like heaven.❤ *one of the best video i have ever watched on calculus on YouTube love from india* 😊❤

[jameshale]

At 7:03, I think the cos(h) should be at the bottom line and the bottom yellow line should not extent your the perimeter of the circle. Correct? So, you’d see that line approach 1. As is, the cos(h) line is already one, since this is a unit circle and it extends from the center to the perimeter.

[stevegerard]

Thank you for taking the time to show the 'how' and 'why' rather than just providing information about a concept. Now it makes sense.

[htg]

Thank you for sharing this beautiful video.
There is a small mistake in minute 07:00, you defined the radius of circle as "cos h", but the "cos h" is part of yellow line which is located inside the right triangle and compose the base of triangle.

[dietrichschoen]

Thank you very much. Your lecture is fantastic! Finally I understud why sin(x)/x=1 is. After hours trying to understand this! Again: Thank you very much.

[geraldramos]

Great video! How did you do the animations?

[tariqmehmoodraza]

hi, awesome video. could you please suggest what apps or software can be used to make such videos. what app are u specifically using for such an interactive geometrical stuff?
regards

[adamlea]

I don't get the explanation of the limit (cos(h)-1)/h. I understand visually that cos(h) -> 1 as h -> 0 but I don't see how it immediately follows that (cos(h)-1)/h -> 0 as h -> 0 since this tends to 0/0 which is undefined.

[jacknisen]

I just wanted to see the formal proof of the cosine limit. You hand waved that.

[avijitdey]

That was wrong. Cosx is the adjacent side in a unit circle. Not the hypotenuse. But in unit circle as angle h->0 we have cosx -> 1 as the radius is 1

[Adler]

Great video, this is how math should be interpreted

[hqs]

5:50 big jum, did you use l'Hopital rule i.e. lim h->0 of sih/h is limit h->0 of cosh/1 (differentiate top and button).

[gentlemandude]

How can the hypotenuse be "cosh" by definition (6:57)? Shouldn't it be equal to sqrt([cosh]^2+[sinh]^2) by Pythagorean theorem? Am I missing something?

[haniperetz]

at 9:02 if H becomes zero does it not mean that sin h divided by h is also zero. since sin(0) = 0 ???

[ss]

@7:30

in a circle ... at any angle, cos h will always be equals to value 1. because the radius of circles is same no matter what is their angle. Please correct me

[erenjager]

I don't understand how the slope = cos(theta)? from minute 1:19

[청동이-up]

continuing speaking out Ah~ Ah~, Thank youCal1fun!

[jackkalver]

This is not proof! Fortunately, I have come up with a proof that is much easier to follow than the whole process of finding the limits and angle-sum identity and then using (sin(x+h)–sin x)/h to derive cos x. It uses the unit circle and the derivative of a parametric function. Someday I may make a video version of it.

[peterbauer]

can you explain more fully when you say later in the video concerning the triangle that "the hypotenuse is cos h by definition"

[peterlohnes]

This is an excellent visualization and helps to understand the results, but doesn't really prove it (which is a very convoluted process than involves proving sin(h)/h is 1 as h approaches zero and (1-cos(h))/h is zero as h approaches zero. It is a very difficult proof and I believe Khan academy shows it well...not easy though