This should be in the calculus section.
Use the sum rule for differentiation.
#\frac{d}{dx}[cos(sin2x)-sin (1/x)]#
#= \frac{d}{dx} cos(sin 2x) - \frac{d}{dx}sin (1/x)#
Focusing on # \frac{d}{dx} cos(sin 2x)#, we use the chain rule. Differentiate the outside function while leaving the inside function inside, then multiply by the derivative of the inside.
The outside function is #cos x#, which has a derivative of #-sin x#
The inside function is #sin 2x#.
For the inside function, the chain rule can be used again. The outside is #sin x# and the inside is #2x# . The derivative of the outside is #cos x# and the derivative of the inside is #2#.
#\frac{d}{dx} cos(sin 2x) = -sin(sin 2x) \cdot (sin 2x)'#
# = -sin(sin 2x) \cdot \cos 2x \cdot 2#
# = -2sin(sin 2x) \cdot \cos 2x#
Focusing on #\frac{d}{dx}sin (1/x)#, we have outside function that is #\sin x#, inside function of #\frac{1}{x} = x^-1#. Derivative of outside is #\cos x#; derivative of inside is #-x^{-2} = -\frac{1}{x^2}#
#\frac{d}{dx}sin (1/x) = \cos(1/x) \cdot (-\frac{1}{x^2}) = -\frac{\cos(1/x)}{x^2}#
Bringing these two sides together:
#\frac{d}{dx}[cos(sin2x)-sin (1/x)]#
#= -2sin(sin 2x) \cdot \cos 2x -[-\frac{\cos(1/x)}{x^2}]#
#= -2sin(sin 2x) \cdot \cos 2x +\frac{\cos(1/x)}{x^2}#
#=\frac{\cos(1/x)}{x^2} -2sin(sin 2x) \cdot \cos 2x#
