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})