Proof of the derivative of sin(x) | Derivatives introduction | AP Calculus AB | Khan Academy
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- Опубликовано: 2 окт 2024
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Proving that the derivative of sin(x) is cos(x).
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"Let's see if I can draw a relatively straight line"
*draws perfectly straight
because he's using an app that can automatically draw straight lines
@@vknl99 no he is just a legend.
@Braylon Brennan It's paid you spammer
@@vknl99 He just locked the cursor for y axis
@@ruzzcraze1862 if he were a legend, he must have drawn a "relatively" straight line
I have not understand anything
That's your fault
Its none of ur business
😂😂😂@@Zolnyx
Skill issue
Pause and ponder.
I love this guys voice and the way he accentuates on words. He just makes anything sound so interesting
The Brazil version is goku voice
No you're telling lie.
@@motiwaledeepak6530 what...??
@@motiwaledeepak6530 ...h.. He.. H.... Hi. Ghbhhhb.
idk but it's kinda hot 🚶♀️‼️
Please can you link the video of the cos x Lim ∆x tends to 0
I mean the one you were talking about cos x and sin x is 1 and 0 respectively
Great proof, Sal! Thanks! I enjoy learning with Khan Academy; it’s a pastime not a sad-time.
Your explanations are amazing 🤩
I always have always felt like I was missing something when trying to learn mathematical concepts.
The “Why” factor...
Today was the day i found the usefulness of proofs.
Any one can learn how to change a light bulb but knowing why to change it is another story and without this key piece of the puzzle we are all left in the dark.
Thank you for the amazing content.
I'm confused, the title of this video is "proof of derivative of sin(x)" and then at 5:01 he says he's not going to do the proof in this video.
No he says tat he won't be proving sinx/x =1....he did derive the proof
Sinx/x can be derived from Euler theorem and talor series
you need the derivative of sinx to proof eulers theorem so you cant use it to proof sindx/dx = 1 since you need the derivative "first".
@@valle2353 it can be found via the squeeze theorem. Sal has already done a video on this
@@valle2353 Also, it's pretty intuitive visually anyway. The smaller an arc is, the better it will resemble a straight line. Naturally, as the limit approaches 0, it eventually becomes a perfect line. Hence lim->0 sinx/x=1
You cannot do this because lim sinx/x =1 based on taylor expansion around zero which relies on knowing the derivative in the first place!
Would like to ask as to whered you get the "cosx+sin∆x+sinxcos∆x" from sin(x+∆x)?
Myrrh Cast it's just a formula you have to know. Search up sin angle addition or something. Doesn't come up tooo often, but very useful
Also it's cosx · sindx, not +
he got if from the trigo. iden. sum & diff. of 2 angles
sin(A+B) = sinAcosB + cosAsinB
@@a1exanderparra thaankksss.
. I rarely see it.
@@a1exanderparra formula you have to understand
There's easier way to prove this;
The derivative of sin(x) = lim ((sin(z)-sin(x))/(z-x)), z->x
Using the equation : sin(a)-sin(b) = 2 cos((a+b)/2) sin((a-b)/2) we get :
lim ((2 cos((z+x)/2) sin((z-x)/2))/(z-x)), z->x
the limit of 2 cos((z+x)/2) as z->x is 2 cos (x)
the limit of sin((z-x)/2)/(z-x) as z->x can be solved by substitution; y = (z-x), when (z->x) (y->0) so we get :
lim sin(y/2)/y as y->0 then we substitute g = y/2 so y = 2g so we get :
lim sin(g)/2g as g->0 = 1/2 lim sin(g)/g as g->0 = 1/2
So finally we get (1/2)*2cos(x) = cos(x)
hope this was clear
actually i never saw this kind of proof. Why can you use z instead of dx
Dx=x-xo or in this case x-z@@nagys36snn
Would have been nice to see the entire proof. Especially the part that is useful in the proof. I know how to manipulate the expressions to get it to look like that but how do i show the lim =1 and 0?
This page has RUclips links to the proofs you mentioned:
www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/modal/a/proving-the-derivatives-of-sinx-and-cosx
what crosshair do you use?
how am I only the second person to like this??
he paid for red dot sight
How does he single out the cosx like that and put it in front of lim
x is a different variable from Δx. Since limiting Δx to 0 doesn't change the value of x, sin x doesn't need to be in the limit. This is just to separate it to create a limit that we know the answer to.
@@chrispyexe its like cos(x) is a constant (k) to the limit
Awesome
I like just using exponents. We know the d/dx of exp(x), so we can just write sin(x) as (exp(ix) - exp(-ix) )/(2i) and it's trivial from there.
But I know of course the proofs done here are meant to avoid complex numbers. Sometimes complex numbers make things simple though :D
So if the derivative of sin(x)=cos(x) does that means if your trying to find the derivative of sin(x) at x=a , f'(a) by plugging in a for x in cos(x)
what
it is incorrect to claim that the limit of sin(Δx ) / Δx as Δx = cos(x) as Δx approaches 0. Here is why:
If sin(Δx)/ Δx = 1 just because you have made some mathematical manipulation with trig-formulas to arrive at the inequality that
1>/ = sin(Δx)/ Δx >/= cos(Δx), and as Δx---> 0, 1= sin(Δx) = cos(0)=1, and basing yourself on the "squeeze" theorem, you jump to the conclusion that sin(Δx)/Δx must also = 1, it'll mean that sin(Δx) = Δx. But, you will never eve be able to prove it in any way. Let's see why you guys have made a fundamental mistake here.
To avoid making it confusing, let Δx = θ . According to the trig-graph of the unit circle, R = 1, y is the side of the right triangle and parallel to the Y axis of the unit circle, and the arc- length subtended by the 2 radii which create the angle θ is S= θR ---> θ = S/R, sin(θ) = y/R
Limit of sin(θ)/θ = Lim {( y/R )/ S/R )= y/S , as θ approaches 0. There is no drawing function here for me to draw the graph. If you draw the graph yourself, you'll see that y is perpendicular to the X axis, and always smaller than the arc-length S because y is also the perpendicular of another smaller triangle of which the base is on the positive X axis , and the positive X axis = radius R of the circle, and the hypotenuse is the segment connecting the 2 ends of the arc S. So, the hypotenuse is greater than the perpendicular y, while it is smaller than the arc-length S. it means that the perpendicular y in sin(θ) = y/R is smaller than the arc-length S.
Since y is always < S, y/S is always < 1---> As θ approaches 0, Lim (sin(θ)/ θ) = y/S is always smaller than 1. Substituting
Limit of sin(θ)/θ i= < 1( smaller than 1) as θ approaches 0, which is the same as substituting Limit of sin(Δx) / Δx) =
Why is he not using h instead… what a weirdo l
Which program do you use to draw on?
good
What would this mean intuitively, does it mean that the tangent of a cosine function at different points varies as a negative sine function???
Yeah, it means that the slope of the tangent of a cosine function is the negative sine function
Wait a minute. You cannot just arbitrarily multiply something by -1. That isn't allowed. Are you muliplying the sin and the (cosx-1) each by -1? Because that would be ok, -1 X -1 = 1. Why did you elide that step?
He is multiplying (cos dx - 1) by -1 to get (1 - cos dx), and to compensate for this, he multiplies again by -1 to get that minus sign in front of the parenthesis. So all in all, he multiplies that term by -1 twice :)
How does the squeeze theorem come into this? i.e. why does lim as dx --> 0 x sin dx/dx = 1?
nvm it's in previous video
Thank you
Thanks, great proof. My math book was super confusing when it came to this subject!
Aaron Georgeson দক্সগক্সক্সক্সক্স
যক্সক্সক্সক্স
এক্সক্সক্সক্স
Hi
Finally here
I went here because of 2-dimensional physics' finding the optimal angle. And I am very, very confused.
So. Did u pass
Thank you
5:15 bruh just use l'hopital rule
What if he factor out sinx first then cosx?
Omg I’m first comment on a khan academy video! So happy! I love your videos btw
For those of you who want to know the proof, it's not explained in this video.
You mention that in other videos you will complete the proof. In which of your videos do you show the complete proof. You might provide a reference number or name for a videos that are used in subsequent subjects.
I found it under the Squeeze Theorem topic. Thanks.
good call though
Thanks a lot for this amazing proof
Playing with such crosshair hopefully make me good at math some day
Thanks a lot
Are you using small angle approximations at the end?
xd
underwhelming