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

  • Thread starter winston2020
  • Start date
In summary: Anyway, you're right winston2020. In summary, the derivative of sin^2(3x) using the chain rule and product rule is 3sin(6x).
  • #1
winston2020
35
0
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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  • #2
nope!

what does chain rule say?

[tex] \frac{d}{dx}sin^2(3x)=2sin(3x)*3=6sin3x[/tex]
 
  • #3
sutupidmath said:
nope!

what does chain rule say?

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

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:

[tex]
= 2sin(3x)*cos(3x)*3
[/tex]
[tex]
= 6* sin(3x)cos(3x)
[/tex]
[tex]
= 3( 2sin(3x)cos(3x) )
[/tex]
[tex]
= 3( sin(2 *3x) )
[/tex]
[tex]
= 3sin(6x)
[/tex]
 
Last edited:
  • #4
chain rule: [f(g(x))]'=f'(g(x))g'(x)
 
  • #5
sutupidmath said:
chain rule: [f(g(x))]'=f'(g(x))g'(x)

Exactly. So then this is my logic:

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

Let [tex] u = 3x [/tex]

And let [tex] v = sin(u) [/tex]

i.e.

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

[tex]
= 2(sin(3x)) * cos(3x) * 3
[/tex]

[tex]
= 6(sin(3x)cos(3x))
[/tex]

And since [tex] 2sin(x)cos(x) = sin(2x) [/tex]

[tex]
= 3( 2sin(3x)cos(3x) )
[/tex]
[tex]
= 3( sin(2*3x) )
[/tex]
[tex]
= 3sin(6x)
[/tex]

Is that not correct?
 
  • #6
That's completely right^^
 
  • #7
Feldoh said:
That's completely right^^

Thank you :smile:
 
  • #8
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).
 
  • #9
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!
 

1. What is the chain rule?

The chain rule is a formula used to find the derivative of a composite function. It states that the derivative of the outer function multiplied by the derivative of the inner function.

2. How is the chain rule applied to sin^2[3x]?

In this case, the outer function is sin^2x and the inner function is 3x. To find the derivative, we first take the derivative of the outer function, which is 2sinx*cosx. Then we multiply it by the derivative of the inner function, which is 3. This results in a final answer of 6sinx*cosx.

3. What is the product rule?

The product rule is a formula used to find the derivative of two functions that are multiplied together. It states that the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.

4. How is the product rule applied to sin^2[3x]?

In this case, the two functions being multiplied together are sinx and sinx. To find the derivative, we use the product rule formula, resulting in a final answer of 2sinx*cosx + 2sin^2x.

5. What is the difference between the chain rule and the product rule?

The chain rule is used when finding the derivative of a composite function, where the outer function is multiplied by the derivative of the inner function. The product rule is used when finding the derivative of two functions multiplied together, where the first function is multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function.

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