Eddie, I'm a 57 year old professional engineer and I still love watching and "relearning" from your videos. Your enthusiasm is amazing. Good for you man and Thank You!!
I am student from India. I have great passion for Mathematics. Sir! Your videos made me feel very much interesting in Mathematics. Sir, Here in India, Teachers don't teach that good. They make us to just memorize the formulas and apply them without any conscience of what we are doing. Thank you very much sir, for your support through your videos. One request! Can you please name me a book that encompasses all topics of Mathematics? I respect you so much sir Thank you
Hi Eddie. I started tutoring math recently and I realized today that I didn't have a great derivation of d/dx{sin(x)} = cos(x) at the high school level (I think of sin and cos as being defined by their Taylor series which makes finding their derivatives trivial, but this introduces other complications). I got as far as the limits of sin(h)/h and (cos(h)-1)/h, and then all I could come up with was the following argument: lim_{h->0} sin(h)/h is the derivative of sin(x) at x=0, and if we imagine traveling around the unit circle at 1 unit per second then we see that when our angle is 0 we are pointing straight up. So if we're traveling straight up at a rate of 1 then our vertical position (i.e. sin(x)) is changing at a rate of 1, and the limit is 1. Anyway, I looked up your videos to see if there was a more rigorous but still accessible way. You really kept up the suspense with those incomplete proofs, but when you finally established sin(x)
I don't think you need to return to first principles to deduce the derivative of cosine. Once you have the derivative of sine, as well as the chain rule, power rule, and constant multiplier rule as established facts, you can determine the derivative of cosine. You can use trig identities to rewrite cosine in terms of sine. Recall that co in cosine means "of the complimentary angle": cos(x) = sin(pi/2 - x) Then, you use the chain rule (which has its own proof from first principles) to take the derivative of sin(pi/2 - x): d/dx sin(pi/2 - x) Let w = pi/2 - x. Thus: dw/dx = -1 d/dx sin(w) = dw/dx * d/dw sin(w) d/dw sin(w) = cos(w) d/dx sin(w) = -1*cos(w) d/dx sin(pi/2 - x) = -1*cos(pi/2 - x) Recall the original trig identity: cos(x) = sin(pi/2 - x) Take complimentary angles of both arguments: cos(pi/2 - x) = sin(pi/2 - (pi/2 - x)) cos(pi/2 - x) = sin(x) And we have our conclusion: d/dx cos(x) = -sin(x)
U mentioned that the graph proof for the derivative of sinx had alot of "holes" in it. How would you or a mathematician go about disproving and pointing out the holes?
I was shown this proof in class and i was sitting there going "That cant be right". I didnt feel right accepting this as a proof for the derivative of sine
@@LOLxUnique Holes in this context, doesn't mean removable zeros. It just means that it isn't a formal proof that guarantees sine's derivative is cosine for all inputs. It shows that sine's derivative generally follows the trend of cosine, but doesn't show how we know it is exactly cosine, and not some other periodic function with the same period and phase.
![Derivatives of Trig Functions (Sin, Cos, Tan) in Calculus - [1-4]](http://i.ytimg.com/vi/ocuZkDGdy1Y/mqdefault.jpg)
Eddie, I'm a 57 year old professional engineer and I still love watching and "relearning" from your videos. Your enthusiasm is amazing. Good for you man and Thank You!!
I am student from India.
I have great passion for Mathematics.
Sir!
Your videos made me feel very much interesting in Mathematics.
Sir, Here in India, Teachers don't teach that good.
They make us to just memorize the formulas and apply them without any conscience of what we are doing.
Thank you very much sir, for your support through your videos.
One request!
Can you please name me a book that encompasses all topics of Mathematics?
I respect you so much sir
Thank you
Hi Eddie. I started tutoring math recently and I realized today that I didn't have a great derivation of d/dx{sin(x)} = cos(x) at the high school level (I think of sin and cos as being defined by their Taylor series which makes finding their derivatives trivial, but this introduces other complications). I got as far as the limits of sin(h)/h and (cos(h)-1)/h, and then all I could come up with was the following argument: lim_{h->0} sin(h)/h is the derivative of sin(x) at x=0, and if we imagine traveling around the unit circle at 1 unit per second then we see that when our angle is 0 we are pointing straight up. So if we're traveling straight up at a rate of 1 then our vertical position (i.e. sin(x)) is changing at a rate of 1, and the limit is 1.
Anyway, I looked up your videos to see if there was a more rigorous but still accessible way. You really kept up the suspense with those incomplete proofs, but when you finally established sin(x)
I don't think you need to return to first principles to deduce the derivative of cosine. Once you have the derivative of sine, as well as the chain rule, power rule, and constant multiplier rule as established facts, you can determine the derivative of cosine.
You can use trig identities to rewrite cosine in terms of sine. Recall that co in cosine means "of the complimentary angle":
cos(x) = sin(pi/2 - x)
Then, you use the chain rule (which has its own proof from first principles) to take the derivative of sin(pi/2 - x):
d/dx sin(pi/2 - x)
Let w = pi/2 - x. Thus:
dw/dx = -1
d/dx sin(w) = dw/dx * d/dw sin(w)
d/dw sin(w) = cos(w)
d/dx sin(w) = -1*cos(w)
d/dx sin(pi/2 - x) = -1*cos(pi/2 - x)
Recall the original trig identity:
cos(x) = sin(pi/2 - x)
Take complimentary angles of both arguments:
cos(pi/2 - x) = sin(pi/2 - (pi/2 - x))
cos(pi/2 - x) = sin(x)
And we have our conclusion:
d/dx cos(x) = -sin(x)
Great tutorial sir.
At 19:32 the inequality should be "≤" because 1< [lim x->0 (x/sin x)]
Your best video
సూపర్
U mentioned that the graph proof for the derivative of sinx had alot of "holes" in it. How would you or a mathematician go about disproving and pointing out the holes?
I was shown this proof in class and i was sitting there going "That cant be right". I didnt feel right accepting this as a proof for the derivative of sine
@@LOLxUnique Holes in this context, doesn't mean removable zeros. It just means that it isn't a formal proof that guarantees sine's derivative is cosine for all inputs. It shows that sine's derivative generally follows the trend of cosine, but doesn't show how we know it is exactly cosine, and not some other periodic function with the same period and phase.
@@carultch haha thanks for the reply many years later
Your first explanation was better. I really don't like it when things get put in place and I have no idea why they are there or where they come from.
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