Question

What is the derivative of `y = sin(sin(sinx))`?

Answer 1

We have `(sin(u))' = u'cos(u)`

where `u = sin(sin(x))`

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

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

Sorry if in your countries you need to use the "chain rules" in France we don't use that, we just remember that :

`(f(g(x)))' = g'(x)*f'(g(x))`