Question

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

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:

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