Differentiating sin(x) from First Principles

[modemanslutning]

Okay, here it is.
First, remember that: sin^2(y) = (1/2)(1-cos(2y))

(cos(h)-1) / h = - (1-cos(h)) / h = - 2sin^2(h/2) / h = - sin(h/2)*sin(h/2) / (h/2)

Just for simplicity, to better see things. Let's call "p = h/2". As h ---> 0, so does p, multiplying by a half here is irrelevant since everything is scaled accordingly. This gives us

lim(p ---> 0): - sin(p)*sin(p)/p

Since the last sin(p)/p = 1 when p goes to zero, that part is known, but we also multiply that with something else, namely sin(p), which goes to zero. So it is essentially - 0 * 1= 0.

And there you go.

DO NOT USE TAYLOR EXPANSION TO SOLVE THIS!

The Taylor expansion is based on the fact that one already knows what the derivatives for sin(x) and cos(x) are (because that is how their expansions are derived). So what you do then is trying to use a tool whose properties are built on derivatives. Essentially trying to prove sin(x)'s derivative, by using sin(x)'s derivative. It is a circle argument.

The same goes for l'Hospital's rule.

Use figures, trigonometric properties, known limits or whatever. Using Taylor or l'Hospital's rule requires knowledge when they can be applied. These are common mistakes on exams.

[lingfai]

in 8:14, lim h is going to zero and cosh -1 is zero, but how can you say that 0/0 =0? Is it something missed?

[taartog]

Eddie Woo, you really have made me interested in Math more than ever.

[vinayseth]

7:55 - BIG MISTAKE there! Yes, limit of cos(h) is approaching 1 as h approaches 0, BUT how can you just neglect the h in the denominator here? This is a big shortcut, with no sensible explanation.

[shanaesung]

As an Aussie student this video was really helpful, especially with the 'in-between' steps that other videos don't discuss.

[harushizuku]

for those who are confuse with cos h - 1/h = 0/0
actually its not 0/0 ^^
see Lets forget about that denominator h
and focus on cos h -1 okay
expand it ( cos h -1 = ( 1- sin^2(h/2) -1) = sin^2(h/2) = ( sin(h/2) x sin(h/2) ) ) i guess u know this
now in sin(y) when y is very small its value = y
example equivalent to
so sin(h/2) [as h -> 0 (very small ) ] = h/2
so that cos h-1/h will become (h/2 x h/2)/h
cancel h and h u will get h/4 put limit u will get 0
i hope its clear to you now ^^
its not 1-1 =0 you just cant forget same variable one over another

[robertcampomizzi]

5:52 Cos looks like a tooth's root when you draw both sides! I'll never forget that now.. I'm healing from dental surgery lmao!!!

[wolf]

sir 1st of all your way of teaching is amazing and i m a big fan of yours .
now about this video, i want u to break lim x-o [sin(x+h)-sin(x)]/h in sina-sinb formula :-
[sin(x+h)-sinx]/h =[ 2cos(x+h+x/2).sin(x+h-x/2)]/h < .symbolises multiplication sign>
=[2cos(2x+h/2).sin(h/2)]/h <assume limit is present as it would be difficult to write it again and again>
now divide and multiply sin(h)/2 by h/2 . then we get :-
=[2cos(2x+h/2) . h/2]/h <as [sin(h/2)]/(h/2)=1>
now, this h/2 divided by h gives 1/2 and = 1/2 . 2cos(2x+h/2) gives ;-
cos(2x+h/2) with limit x-0<x tends to zero> we get
cos(2x/2) = cos(x)

[abdurrahman]

cos h - 1 / h gains 0/0 form when limit is applied, how'd u justify that ?

[eyobembaye]

the best Math teacher i have ever seen

[peterlohnes]

For those of you puzzled, to really prove the limit of 1-cosh/h and sinh/h, you should really NOT do this or memorize what he did, but examine other sources (google Khan academy has a very long winded explanation but its better than making the grand assumptions he did).

[PatricioHondagneuRoig]

Um, Cos(h)-1/h when approaching h=0 gives you an indetermination. I'm quite confused about that. Great video nonetheless!

[-GCET--iuji]

u r gr8.make the topic interesting and u r always full of energy gr8.

[gdaaps]

First principles are sometimes also referred to as definitions

[SercanPy]

So much energy! That was really nice, but I was really hoping you'd go through the Lim h-> 0 sin(h)/h = 1 part. Can you say that sin(h) = h since they both equal zero as h ->0 ?

[sirfadong]

7:45 why you no explain small angle approximations?

[allanmarrance]

I can't stop looking at the speaker's face. He is handsome😍 thanks for the knowledge

[janov]

love you explanations...and your enthusiasm...thank you for sharing...

[thepositionalplay]

what is the age of those students? And in what class do u introduce these topics in ur country?

[chrisofnottingham]

This could have been done without the factoring by sin x and just splitting the numerator into three fractions over h. And from sketching a right angle triangle it is easy to to see that (cos h)/h --> 1 as h --> 0.