Question

let $f(x)=7\sin(x)-2x^{3}$. $f(x)=$

Explanation:

Step1: Differentiate \(7\sin(x)\)

The derivative of \(\sin(x)\) is \(\cos(x)\). Using the constant multiple rule \((cf(x))^\prime = cf^\prime(x)\) (where \(c = 7\) and \(f(x)=\sin(x)\)), the derivative of \(7\sin(x)\) is \(7\cos(x)\).

Step2: Differentiate \(-2x^{3}\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\) (where \(n = 3\) and \(c=-2\)), we have \((-2x^{3})^\prime=-2\times3x^{3 - 1}=-6x^{2}\).

Step3: Find \(f^\prime(x)\)

By the sum - difference rule \((u\pm v)^\prime=u^\prime\pm v^\prime\) (where \(u = 7\sin(x)\) and \(v = 2x^{3}\)), \(f^\prime(x)=(7\sin(x))^\prime-(2x^{3})^\prime\).
Substituting the results from Step1 and Step2, we get \(f^\prime(x)=7\cos(x)-6x^{2}\).

Answer:

\(7\cos(x)-6x^{2}\)