Question

g(x)=2x\sqrt{x}\sin(x)
which sequence of rules can be used in order to differentiate g in its current form?
choose 1 answer
chain rule, then chain rule again
product rule, then product rule again
product rule, then chain rule
chain rule, then product rule

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