f(x)=sin^2 3x
2006-11-28 09:57:11 · 5 answers · asked by julia t 1 in Science & Mathematics ➔ Mathematics
You need to use the chain rule a couple of times.
How do you differentiate (something)^2 ? Make it 2*something. Then multiply by (something) differentiated.
In this case, (sin 3x)^2 differentiates to 2(sin 3x) * (sin 3x differentiated).
Now, how do you differentiate sin(something)? Make it cos(something), then multiply by (something) differentiated.
So sin 3x differentiates to cos 3x * 3.
Overall, you get 2 sin 3x cos 3x * 3 = 6 sin 3x cos 3x.
2006-11-28 10:06:06 · answer #1 · answered by stephen m 4 · 0⤊ 1⤋
2*sin(3x)*cos(3x)*3 which can be simplified using trigonometric identities into 3*sin(6x).
2006-11-28 18:02:08 · answer #2 · answered by The Prince 6 · 0⤊ 1⤋
Chain rule:
f'(x) = d/dx[sin 3x]^2 = 2sin 3x + d/dx[sin 3x]
now d/dx[sin z] = (dz/dx)cos z where z = 3x
d/dx[sin z] = 3cos 3x
and then
f'(x) = 2sin 3x + 3 cos 3x
2006-11-28 18:07:33 · answer #3 · answered by kellenraid 6 · 0⤊ 1⤋
6*sin(3x)*cos(3x)
2006-11-28 17:59:39 · answer #4 · answered by sft2hrdtco 4 · 0⤊ 1⤋
6sin(3)*cos(3)
Ohh good ole Calculus :)
2006-11-28 18:02:59 · answer #5 · answered by Summer_Laine 2 · 0⤊ 1⤋