g(x)=2x\\sqrt{x}\\sin(x)\nwhich sequence of rules can be used in order to differentiate g in its current…

Explanation:

Step1: Analyze the function structure

The function (g(x) = 2x\sqrt{x}\sin(x)) can be seen as a product of (2x) and (\sqrt{x}\sin(x)). So, the product rule ((uv)^\prime=u^\prime v + uv^\prime) (where (u = 2x) and (v=\sqrt{x}\sin(x))) is applicable first.

Step2: Analyze the inner - function

For the function (v=\sqrt{x}\sin(x)) (which is part of the original product), it is also a product of two functions (y_1=\sqrt{x}=x^{\frac{1}{2}}) and (y_2 = \sin(x)). But to differentiate (\sqrt{x}), we use the power rule ((x^n)^\prime=nx^{n - 1}) (a special case related to the chain rule when the outer function is (y = u^n) and (u=x), (y^\prime=ny^{n - 1}\cdot u^\prime), here (u^\prime = 1)). However, if we consider the general form of differentiating a composition of functions (even in simple cases like (y=\sqrt{x})), the concept is related to the chain rule. In the context of differentiating (v=\sqrt{x}\sin(x)) (after the first - level product rule for (g(x))), we can think of differentiating (\sqrt{x}) (using the idea from the chain rule) and then using the product rule for (v). But if we just consider the rules for the given problem's structure: The function (g(x)) is a product of two functions (u = 2x) and (v=\sqrt{x}\sin(x)). To differentiate (g(x)), we first use the product rule (g^\prime(x)=(2x)^\prime\cdot\sqrt{x}\sin(x)+2x\cdot(\sqrt{x}\sin(x))^\prime). Then, to find ((\sqrt{x}\sin(x))^\prime), we use the product rule again (if we consider (\sqrt{x}) as a simple power function, but if we are strict with the rule names in the context of the problem's options, we note that (\sqrt{x}) can be thought of as a composition (y = u^{\frac{1}{2}}) with (u = x) (a trivial chain - rule case) and then multiplied by (\sin(x))). But if we consider the rules as per the options: The function (g(x)) is a product of two functions. Let (a = 2x) and (b=\sqrt{x}\sin(x)). By the product rule (g^\prime(x)=a^\prime b+ab^\prime). Now, for (b=\sqrt{x}\sin(x)), let (m=\sqrt{x}) and (n = \sin(x)). By the product rule (b^\prime=m^\prime n+mn^\prime).

Answer:

C. Product rule, then chain rule