Question

dy/dx of cos^2 3x

Answer 1

use the chain rule
d/dy (y) = d/dy cos²(3x)

1 = d/dy cos²(3x)

let u = cos(3x)
d/du u² = 2u
dx/dy cos(3x) = -3sin(3x) dx/dy

1 = 2 cos(3x) * -3sin(3x) dx/dy
1 = -6 cos(3x) sin(3x) dx/dy
1 = -3 (2cos(3x) sin(3x)) dx/dy
1 = -3 sin(6x) dx/dy
dx/dy = -1/[3sin(6x)]

Answer 2

hmm confused as to y you made it equal to 1

Answer 3

-1/[3sin(6x)]

Answer 4

1 = d/dy cos²(3x)
idk I dont remember making it equal to 1. I kinda see your logic but I dont remember doing that stuff

Answer 5

its ok

Answer 6

The answer should just be -3sin(6x)

Answer 7

You somehow got confused there

Answer 8

@silvercharm how did you get 1/-3sin(6x)?

Answer 9

Looks like a typo that just carried through:
\[\frac{d}{dx} \cos²(3x)\] let u = cos(3x) \[\frac{d}{du} u^2 = 2u\] Substitute back in and you have: \[\frac{dy}{dx} 2[\cos(3x)] \rightarrow 2[(\cos(3x)((3)(-\sin(3x))] \rightarrow 2[(\cos(3x))*-3*(\sin(3x))] \]
Combining for \[cos(3x)(-6sin(3x))\]
since 2 sin(x)cos(x) = sin(2x) you can divide out your 2 and rewrite this to read:\[-6\sin(3x))\cos(3x) \rightarrow \frac{-6}{2}\sin(2*3x))\cos(2*3x) \rightarrow -3\sin(6x)\]