How do you find the derivative of #y=cos(cosx)# ?

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Answer 1 · 2018-04-05T16:55:49

#dy/dx=sinx*sin(cosx)#

First from the differentiation of trigonometric functions :

#d/dx cosx=-sinxdx#

#d/dxcosu=-sinu*du#

Where # u# is a function of #x#

so when You differentiate #y=cos(cosx)#

You get the following:

#dy= -sin(cosx)dcosx#

which gives You:

#dy=(-sin(cosx))*(-sinx*dx)#

and by simpilification you get:

#dy/dxcosx=sinx*sin(cosx)#