$$ \frac { d } { d x } - 4 x ^ { 3 } - sin ( x ) = $$

Explanation:

Step1: Apply the sum/difference rule

The sum/difference rule of differentiation states that ((u - v)'=u' - v'). Let (u = - 4x^{3}) and (v=\sin(x)). So, (\frac{d}{dx}(-4x^{3}-\sin(x))=\frac{d}{dx}(-4x^{3})-\frac{d}{dx}(\sin(x)))

Step2: Differentiate (u = - 4x^{3})

Using the power rule (\frac{d}{dx}(ax^{n})=nax^{n - 1}), for (a=-4) and (n = 3), we have (\frac{d}{dx}(-4x^{3})=-4\times3x^{3-1}=-12x^{2})

Step3: Differentiate (v=\sin(x))

The derivative of (\sin(x)) with respect to (x) is (\cos(x)), so (\frac{d}{dx}(\sin(x))=\cos(x))

Answer:

(-12x^{2}-\cos(x))