What is the derivative of #y = cos(cos(cos(x)))#?

1 Answer
Oct 9, 2016

Answer:

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

Explanation:

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 Chain Rule, 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)#