f(x)=xcos(x)\nhow would you rewrite f(x) so it can be differentiated using the power rule?\nchoose 1…

Explanation:

Step1: Recall the power rule

The power rule is (\frac{d}{dx}(x^n)=nx^{n - 1}), which is for functions of the form (y = x^n) where (n) is a real - number.

Step2: Analyze option A

For (y=\cos(x^{2})), we need to use the chain rule. Let (u = x^{2}), then (y=\cos(u)). By the chain rule (\frac{dy}{dx}=-\sin(u)\cdot2x=- 2x\sin(x^{2})). It is not in a form that can be differentiated using the power rule.

Step3: Analyze option B

First, note that (\cos^{-1}(x)) is the inverse cosine function ((\arccos(x))), and (\frac{x}{\cos^{-1}(x)}) is a quotient of a polynomial and an inverse - trigonometric function. We would use the quotient rule (\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}) (where (u = x), (u'=1) and (v=\arccos(x)), (v'=-\frac{1}{\sqrt{1 - x^{2}}})). It is not in a form for the power rule.

Step4: Analyze option C

(\frac{\cos(x)}{x^{-1}}=x\cos(x)) (since (x^{-1}=\frac{1}{x}) and (\frac{\cos(x)}{x^{-1}}=\cos(x)\div\frac{1}{x}=x\cos(x))). To differentiate (y = x\cos(x)), we use the product rule ((uv)'=u'v+uv') (where (u = x), (u' = 1) and (v=\cos(x)), (v'=-\sin(x))). The product rule is different from the power rule.

Answer:

D. This is not possible.