How was the chain rule used to find the derivative of a trigonometric function?

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Discussion Overview

The discussion centers around the application of the chain rule in calculus, specifically regarding the derivation of trigonometric functions. Participants are trying to clarify the steps involved in transitioning between different forms of a limit expression related to derivatives.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how a limit expression changes from "lim as x -> 0 { cos 3x (dy/dx) - sin 3x (dy/dx) }" to "lim as x -> 0 { 3 cos 3x (dy/dx) - 3 sin 3x (dy/dx)}."
  • Another participant doubts that the transition can be explained without showing all steps and suggests that the multiplication by 3 could imply a relationship between the two limits.
  • One participant asserts that the limits equal "-3 sin 3x," questioning how the derivatives of the trigonometric functions are derived to include the factor of 3.
  • Another participant challenges the claim that the limits equal "-3 sin 3x," emphasizing the importance of the limit notation and the need for the complete context of the problem.
  • A later reply suggests that the application of the chain rule is what allows for the multiplication by 3 in the derivatives of the trigonometric functions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the limit expressions or the steps involved in deriving them. There are competing views on how the expressions relate to each other and what the correct interpretations are.

Contextual Notes

Participants note the absence of the complete expression or equation that is being analyzed, which limits the ability to fully understand the derivation steps. The discussion also highlights the importance of the chain rule in differentiating composite functions, but the specifics remain unclear.

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

lim as x -> 0 { 3 cos 3x (dy/dx) - 3 sin 3x (dy/dx)}

from

lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }

can some one explain pls?
 
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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.
 
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...
 
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?
 
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!
 
PrudensOptimus said:
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...

To get that by getting the derivative of 3x which is 3 then copy cos3x and you will get
"3 cos 3x" same as "sin 3x to be 3 sin 3x". Not equal to -3 sin3x.
 
Last edited:
if i understand what you mean, they just used the chain rule for derivatives.
 

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