Differentiating sin^2[3x]: Chain & Product Rules

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around the differentiation of the function sin²(3x), focusing on the application of the chain and product rules in calculus. Participants explore various methods for arriving at the derivative and clarify their reasoning throughout the process.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests using the product rule by expressing sin²(3x) as (sin(3x))(sin(3x)) and applies the chain and product rules, initially arriving at 3*sin(6x).
  • Another participant challenges this approach, stating that the correct application of the chain rule yields 6*sin(3x)cos(3x) instead.
  • A later reply elaborates on the chain rule, indicating that the derivative of sin(3x) should be included, leading to the expression 6*sin(3x)cos(3x) and ultimately simplifying to 3*sin(6x).
  • Several participants confirm the correctness of the final expression 3*sin(6x) while discussing the steps involved in the differentiation process.
  • One participant expresses confusion about the application of the chain rule and the role of cosine in the differentiation process, prompting further clarification from others.

Areas of Agreement / Disagreement

While some participants agree on the final result of 3*sin(6x), there is initial disagreement on the method of differentiation and the application of the chain rule versus the product rule. The discussion reflects varying levels of confidence and understanding among participants.

Contextual Notes

Participants express uncertainty about the steps involved in differentiation, particularly regarding the application of the chain rule and the treatment of the inner function sin(3x). There is also a lack of consensus on the initial approach to the problem.

Who May Find This Useful

This discussion may be useful for students or individuals seeking clarification on the differentiation of composite functions, particularly in the context of trigonometric identities and rules of calculus.

winston2020
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I'm not sure how to differentiate sin^2[3x]. Although, I think it's just d/dx( (sin[3x])(sin[3x]) ). So, just chain and product rules should do it. Is that right?

EDIT: I've followed through with the above method, and I got 3*sin(6x). Is that correct?
 
Last edited:
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nope!

what does chain rule say?

\frac{d}{dx}sin^2(3x)=2sin(3x)*3=6sin3x
 
sutupidmath said:
nope!

what does chain rule say?

\frac{d}{dx}sin^2(3x)=2sin(3x)*3=6sin3x

Oh man, I really over complicated things. Thanks :D

EDIT: Wait though... shouldn't sin switch to cos at some point?

Doesn't the chain rule mean it should go something like this:

<br /> = 2sin(3x)*cos(3x)*3<br />
<br /> = 6* sin(3x)cos(3x)<br />
<br /> = 3( 2sin(3x)cos(3x) )<br />
<br /> = 3( sin(2 *3x) )<br />
<br /> = 3sin(6x)<br />
 
Last edited:
chain rule: [f(g(x))]'=f'(g(x))g'(x)
 
sutupidmath said:
chain rule: [f(g(x))]'=f'(g(x))g'(x)

Exactly. So then this is my logic:

Since sin^2(u) = [sin(u)]^2

Let u = 3x

And let v = sin(u)

i.e.

\frac{d}{dx}v^2 = 2v = 2(sin(u)) * \frac{d}{dx}sin(u) = 2(sin(3x)) * cos(3x) * \frac{d}{dx}3x

<br /> = 2(sin(3x)) * cos(3x) * 3<br />

<br /> = 6(sin(3x)cos(3x))<br />

And since 2sin(x)cos(x) = sin(2x)

<br /> = 3( 2sin(3x)cos(3x) )<br />
<br /> = 3( sin(2*3x) )<br />
<br /> = 3sin(6x)<br />

Is that not correct?
 
That's completely right^^
 
Feldoh said:
That's completely right^^

Thank you :smile:
 
I believe winston2020 is correct.

Sutupidmath, the g(x) function is sin(3x) so you have to take the derivative of that, which is 3cos(3x), not the derivative of 3x.

winston, you should be a bit more confident in your answers :). You did mention that you could do this via chain rule or product rule, so that gives you a way to check your answer. The idea is not to worry about 3x in on (sin(3x))2. Use the well known rule for dealing with powers, then take the derivative of the "inside" function, sin(3x), which is just 3cos(3x) and multiply to get 2sin(3x)3cos(3x) = 6sin(3x)cos(3x) = 3*2sin(3x)cos(3x) = 3sin(6x).
 
snipez90 said:
I believe winston2020 is correct.

Sutupidmath, the g(x) function is sin(3x) so you have to take the derivative of that, which is 3cos(3x), not the derivative of 3x.

.

My bad lol..i don't know what i was thinking!
 

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