Derivative of y=sin[cos(sinx)] ?

Question

Hi, can you please show me steps on how to do that ^

Answer 1

Differentiating this function involves use of the chain rule.  

 

Recall that if u is a function of x, then (d/dx)(cosu) = -sinu(du/dx)

 

                                                      (d/dx)(sinu) = cosu(du/dx)

 

So, if y = sin[cos(sinx)],

 

      then dy/dx = cos[cos(sinx)] (d/dx)[cos(sinx)]

 

                      = cos[cos(sinx)][-sin(sinx)(d/dx)(sinx)]

 

                      = cos[cos(sinx)](-sin(sinx))(cosx)

 

                      = -cosx(sin(sinx))cos[cos(sinx)]