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Derivatives of Trig functions 
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#1
Jun403, 09:31 PM

P: 640

I didn't understand how they arrived at this step:



#2
Jun503, 06:42 AM

Math
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PF Gold
P: 39,363

I doubt that anyone can explain how one step follows from the other because you haven't shown the whole steps.
I assume what you have says that "lim as x > 0 { 3 cos 3x (dy/dx)  3 sin 3x (dy/dx)}" EQUALS something and that that follows from the fact that "lim as x > 0 { cos 3x (dy/dx)  sin3x (dy/dx) }" equals something else. Unfortunately, you haven't told us what they equal. Since the only difference between the two that I can see is that the first above is multiplied by 3, I would suspect that "lim as x > 0 { 3cos 3x (dy/dx)  3sin3x (dy/dx) }= 3L" follows from "lim as x > 0 { cos 3x (dy/dx)  sin3x (dy/dx) }= L" by multiplying the equation by 3. But of course, I can't be sure. 


#3
Jun503, 07:53 AM

P: 640

they equal to 3 sin3x.
I was just wondering how did they get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x... 


#4
Jun503, 09:34 AM

P: 1,047

Derivatives of Trig functions
Please tell us exactly what it is that you (or "they") are trying to prove  not just this one step, but the overall context.
And, what is the entire expression or equation you are starting out with? 


#5
Jun503, 01:52 PM

Math
Emeritus
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PF Gold
P: 39,363

No, they DON'T "equal to 3 sin3x." They can't because each has a
"lim as x>0". Once again, I can't say precisely what happened because you still haven't given us the whole thing but the obvious way to "get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x" is to multiply by 3! 


#6
Sep607, 11:53 AM

P: 1

"3 cos 3x" same as "sin 3x to be 3 sin 3x". Not equal to 3 sin3x. 


#7
Feb2008, 04:21 PM

P: 29

if i understand what you mean, they just used the chain rule for derivatives.



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