d/dx (sin^2[3x])

winston2020

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?

sutupidmath

nope!

what does chain rule say?

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

winston2020

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]

sutupidmath

chain rule: [f(g(x))]'=f'(g(x))g'(x)

winston2020

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?

Feldoh

That's completely right^^

winston2020

Thank you

snipez90

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).

sutupidmath

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