Question

Could somebody help me to derivate this
g(w)= sin (cos 3w)- 3 cos^2 2w

Answer 1

cos ( cos (3w))*(-sin(3w))*3-

Answer 2

6cos(2w)*(-sin(2w))*2

Answer 3

-3 cos (cos (3w)) sin(3w)+12cos(2w)sin(2w)

Answer 4

* sin (2x)=2 sin x cos
-3 cos (cos (3w)) sin(3w)+6 sin(4w)

Answer 5

do each term separately and then add the up at the end.

taking them in turn:

: g = sin (cos 3w)
you might wish to break this up a bit more to make it easier.
you could say that g = sin (x) where x = cos (3w)
then use chain rule: dg/dw = dg/dx * dx/dw


: g = - 3 cos^2 2w
this is easier but again, if you find it daunting, you might try doing the same kind of thing.
you can say g = - 3 x^2, where x = cos (2w)
chain rule it: dg/dw = dg/dx * dx/dw

Answer 6

@IrishBoy123 i don't get very well could explain me how to solve g=-3cos ^2 2w

Answer 7

if g=-3cos ^2 2w

then let x = cos 2w

thus g = -3 x^2

yes?

Answer 8

HI!!

Answer 9

what are you looking for the derivative of?

Answer 10

it is this?
\[g(w)= \sin (cos (3w))- 3 \cos^2 (2w)\]

Answer 11

yeah

Answer 12

gotta use the chain rule all the way down here

Answer 13

first lets look at
\[\sin(\cos(3w))\] the derivative of sine is cosine so start with
\[\cos(\cos(3w)\times \] then the derivative of cosine is minus sine so
\[\cos(\cos(3w)\times (-\sin(3w))\times \]

Answer 14

and finally the derivative of \(3w\) is \(3\) final answer \[\cos(\cos(3w)\times (-\sin(3w))\times3\]

Answer 15

clean it up a little and write
\[-3\cos(\cos(3w))\sin(3w)\] and is just the first part

Answer 16

you good with that? we still have to do
\[-3\cos^2(2w)\]

Answer 17

In that part is where I got confuse because is square

Answer 18

\[\frac{d}{dx}f^2(x)=2f(x)f'(x)\]

Answer 19

the derivative of something squared is two time something, times the derivative of something

Answer 20

in your case the something is \(\cos(2w)\) and the derivative of that is
\[2\sin(2w)\]

Answer 21

the derivative of the second part is therefore
\[+12\times \cos(2w)\sin(2w)\]

Answer 22

then whats is next S:?