When writing #cos(x)# is is important to include the #x# and not just write #cos#. Assuming you meant the former, we shall prove that
if #y = sin(x)sin(x) - cos(x)cos(x)#
then #dy/dt = 2sin(2x)#.
Since the sinus in the derivative has #2x# as an argument, rather than just #x#, this is a sign that we have to use a "double angle formula". Either looking in a table, or remembering, we find that
#cos(2x) = cos^2(x) - sin^2(x)#.
Knowing this, we can rewrite #y# as
#y = sin^2(x) - cos^2(x)#
#y = -(cos^2(x) - sin^2(x))#
#y = -cos(2x)#.
Now, taking #g(x) = 2x# as an inner function and #y(g(x)) = -cos(g(x))# as an outer function, we get from the Chain rule that
#dy/dx = (dy)/(dg) *(dg)/(dx)#
#dy/dx= -(-sin(g(x))) *2#
#dy/dx = 2sin(2x)#,
since #d/(dg) cos (g) = -sin(g)# and #d/(dx) 2x = 2#. Thereby you have showed what you sought out to show.
Note: You could also solve the problem by taking the derivative of #y# with respect to #x#, and only later use a double angle formula. Here, as you probably noticed, we first used a double angle formula and then took the derivative.
