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
