Question

What is the derivative of `sin(sin(cos(x)sin(x)))`?

Answer 1

`=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))cos (2x)`

Explanation:

I assume that x is in radian measure so that

`(sin x)'=cos x and (cos x)'=-sin x`.

Applying chain rule,

`(sin(sin(cos(x)sin(x)))'`

`=cos(sin(cos(x)sin(x)))(sin((cos(x)sin(x))'`

`=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))(cos(x)(sin(x))'+(cos(x))'sin(x))`

`=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))(cos^2(x)-sin^2(x))`

`=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))cos (2x)`