Differentiation of sin function where's my mistake?

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Discussion Overview

The discussion revolves around the differentiation of the sine function, particularly focusing on the application of the chain rule in the context of differentiating sin(3x). Participants are examining a specific mistake in the differentiation process and clarifying the correct notation and reasoning involved.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a differentiation attempt, stating that dy/dx sin(3x) = 3cos(3x) and expresses confusion over a resulting contradiction.
  • Another participant points out that the notation "dy/dx sin(x)" is not well-defined and suggests using d/dx instead, which leads to a clearer understanding of the chain rule application.
  • Some participants emphasize the importance of the chain rule, noting that it is necessary for correctly differentiating sin(3x) and that the leading factor of 3 arises from this rule.
  • There is a discussion about the difference between the notation for derivatives as a noun (dy/dx) versus an operator (d/dx), with some participants clarifying this distinction.
  • One participant admits to relying on a table of derivatives without fully understanding the chain rule, indicating a gap in their comprehension.

Areas of Agreement / Disagreement

Participants generally agree on the need for proper notation and the application of the chain rule, but there is no consensus on the initial mistake made in the differentiation process. Some participants express confusion and seek clarification, indicating that the discussion remains unresolved.

Contextual Notes

There are limitations in the clarity of notation and understanding of differentiation rules among participants, particularly regarding the chain rule and its application in this context.

Hawksteinman
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I was thinking and came up with this. I know it's wrong but can't find the mistake :(

dy/dx sin(x) = cos(x)
dy/dx sin(kx) = kcos(kx)

So dy/dx sin(3x) = 3cos(3x)
Now let Y = 3x
dy/dx sin(Y) = cos(Y) = cos(3x)

3cos(3x) = cos(3x)
3 = 1

Where is the mistake?
 
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Hawksteinman said:
I was thinking and came up with this. I know it's wrong but can't find the mistake :(

dy/dx sin(x) = cos(x)
dy/dx sin(kx) = kcos(kx)

So dy/dx sin(3x) = 3cos(3x)
Now let Y = 3x
dy/dx sin(Y) = cos(Y) = cos(3x)

3cos(3x) = cos(3x)
3 = 1

Where is the mistake?
## \dfrac{d}{dx}\sin(Y) \neq \cos(Y)##
You've made the chain rule vanish by pretending ##Y=y##.
 
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Your notation doesn’t make sense.
dy/dx is the derivative of y with respect to x. “dy/dx sin(x)” is not a well-defined expression.

What you nean is d/dx sin(x). And suddenly the issue disappears:

d/dx sin(3x)=3cos(3x)
d/dx sin(y)=? - here we need the chain rule and d/dx 3x = 3:
d/dx sin(y) = cos(y) d/dx y = cos(y) * 3 = 3cos(3x)
 
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mfb said:
Your notation doesn’t make sense.
dy/dx is the derivative of y with respect to x. “dy/dx sin(x)” is not a well-defined expression.

What you nean is d/dx sin(x). And suddenly the issue disappears:

d/dx sin(3x)=3cos(x)
d/dx sin(y)=? - here we need the chain rule and d/dx 3x = 3:
d/dx sin(y) = cos(y) d/dx y = cos(y) * 3

I haven't done the chain rule yet I'll need to look into that :)
 
Basically:
If y =3x changes three times as fast as x, and sin(y) changes cos(y) times as fast as y, then sin(y)=sin(3x) changes 3*cos(y) = 3*cos(3x) times as fast as x.
 
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Hawksteinman said:
So dy/dx sin(3x) = 3cos(3x)
As already noted, the above should be ##\frac d {dx}\left(\sin(3x)\right) = 3\cos(3x)##
##\frac{dy}{dx}## is the derivative of y with respect to x, so it is a thing, a noun.
##\frac d{dx}## is the operator that signifies taking the derivative of something with respect to x. It is an action that hasn't completed yet, a verb. Don't confuse these two things.

Hawksteinman said:
I haven't done the chain rule yet I'll need to look into that :)
Actually, you used the chain rule in the first line of what I quoted, above. The chain rule is what gives you that leading factor of 3.
 
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Mark44 said:
As already noted, the above should be ##\frac d {dx}\left(\sin(3x)\right) = 3\cos(3x)##
##\frac{dy}{dx}## is the derivative of y with respect to x, so it is a thing, a noun.
##\frac d{dx}## is the operator that signifies taking the derivative of something with respect to x. It is an action that hasn't completed yet, a verb. Don't confuse these two things.

Actually, you used the chain rule in the first line of what I quoted, above. The chain rule is what gives you that leading factor of 3.

I don't know, I just used a table of derivatives :/
 
Hawksteinman said:
I don't know, I just used a table of derivatives :/
They were using the chain rule in the table.
 

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