Question

let ( g(x)=-2cos(x)+3x^{3} ). ( g(x)= )

Explanation:

Step1: Differentiate \(-2\cos(x)\)

The derivative of \(\cos(x)\) is \(-\sin(x)\). Using the constant multiple rule, the derivative of \(-2\cos(x)\) is \(-2\times(-\sin(x)) = 2\sin(x)\).

Step2: Differentiate \(3x^{3}\)

Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), for \(n = 3\) and the constant multiple rule, the derivative of \(3x^{3}\) is \(3\times3x^{3-1}=9x^{2}\).

Step3: Sum the derivatives

By the sum rule of differentiation \((u + v)^\prime=u^\prime+v^\prime\), where \(u=-2\cos(x)\) and \(v = 3x^{3}\), we have \(g^\prime(x)=2\sin(x)+9x^{2}\).

Answer:

\(2\sin(x)+9x^{2}\)