Question

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

Answer 1

By the chain rule `f'(x)=cos(xcosx)*(cosx-xsinx)`

Explanation:

`(df)/dx=(d(sin(g(x))))/dx=cos(g(x))*(dg(x))/dx`
in our case `g(x)=x*cosx` so
`g'(x)=cosx-xsinx`