Question

What is the derivative of `sin(cosx)`?

Answer 1

`dy/dx=-sinxcos(cosx)`

Explanation:

You need to use the chain rule,
`d/dxf(g(x)) =f'(g(x))g'(x)` or,`dy/dx=dy/(du)(du)/dx`

So, if `y=sin(cosx) => dy/dx=cos(cosx)d/dxcosx`
`:, dy/dx=cos(cosx)(-sinx)`
`:. dy/dx=-sinxcos(cosx)`