##(dy)/(dx)=-sin(cos(cos(x)))sin(cosx)sin(x)##

In order to differentiate a function of a function, say ##y, =f(g(x))##, where we have to find ##(dy)/(dx)##, we need to do (a) substitute ##u=g(x)##, which gives us ##y=f(u)##. Then we need to use a formula called , which states that ##(dy)/(dx)=(dy)/(du)xx(du)/(dx)##. In fact if we have something like ##y=f(g(h(x)))##, we can have ##(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)##

Hence for ##y=cos(cos(cos(x)))##

##(dy)/(dx)=-sin(cos(cos(x)))xxd/(dx)(cos(cos(x)))##

= ##-sin(cos(cos(x)))xx(-sin(cosx))xxd/(dx)cos(x)##

= ##-sin(cos(cos(x)))xx(-sin(cosx))xx-sin(x)##

= ##-sin(cos(cos(x)))sin(cosx)sin(x)##

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