Question

How do you find the derivative of `sin(cos(6x))`?

Answer 1

`(dy)/(dx)=-6cos(cos(6x))sin(6x)`

Explanation:

We use the chain rule a bunch.

`y=sin(u)`, `u=cos(v)`, `v=6x`.

The chain rule says:
`dy/dx=dy/(du)\*(du)/(dv)\*(dv)/(dx)`

`dy/(du) = cos(u) = cos(cos(v))=cos(cos(6x))`
`(du)/(dv)=-sin(v)=-sin(6x)`
`(dv)/(dx)=6`

So,

`(dy)/(dx)=cos(cos(6x))\*(-sin(6x))\*6`

cleaning up:

`(dy)/(dx)=-6cos(cos(6x))sin(6x)`

Answer 2

`-6cos(cos(6x))sin(6x)`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(h(x)))" then"`

`dy/dx=f'(g(h(x)))xxg'(h(x))xxg'(x)`

`rArrd/dx(sin(cos(6x))`

`=cos(cos(6x))xxd/dx(cos(6x))xxd/dx(6x)`

`=cos(cos(6x))(-sin(6x))(6)`

`=-6cos(cos(6x))sin(6x)`