Question

The derivative of cos^2(4x) is 2cos(4x)(-sin(4x))(4)
Why do I take the derivative of the 4x? Why is that last 4 present?

Answer 1

that's how the chain rule works, u have to derive from inner till outer

Answer 2

derivative of 4x is 4

Answer 3

derivative of (...)^2 = 2(...)

Answer 4

derivative of cos() = -sin()

Answer 5

times up of them

Answer 6

Yes, I understand all of the rest of it. But I don't know why I have to take the derivative of the 4x. I guess I just don't understand the chain rule.

Answer 7

well, i will show step by step

Answer 8

we have
y = cos^2(4x)

let u = 4x
du/dx = 4, agree ?

Answer 9

yes

Answer 10

ok, now we have
y = cos^2 (u)

let v = cos(u)
dv/du = -sin(u), agree ?

Answer 11

ok

Answer 12

finally, now we have
y = v^2
dy/dv = 2v, right ?

Answer 13

right

Answer 14

thus, to get y' = dy/dx just multiply of them
dy/dx = du/dx * dv/du * dy/dv
dy/dx = 4 * (-sin(u)) * 2v

Answer 15

but, the answer must be in x term
so, subtitute back that u = 4x and
v = cos(u) = cos(4x)

Answer 16

therefore,
dy/dx = 4 * (-sin(u)) * 2v
dy/dx = 4 * (-sin(4x)) * 2cos(4x)

Answer 17

that's look too long way, so u have to doing the chain rule directly to be easier

Answer 18

Ok, so cos(x) would actually have a derivative of
-sin(x) * 1
where the one is the derivative of x?

Answer 19

exactly, yes!

Answer 20

Thank you very much! You've helped a lot

Answer 21

my pleasure could help u :)