let ( h(x)=x cos ^{3}(x) ).\nfind ( h^{prime}(x) ).\nchoose 1 answer:\n(a) ( cos ^{2}(x)(cos (x)+3 x) )\n(b)…

Explanation:

Step1: Apply the product rule

The product rule states that if (h(x)=u(x)v(x)), then (h'(x)=u'(x)v(x)+u(x)v'(x)). Here, (u(x) = x) and (v(x)=\cos^{3}(x)). The derivative of (u(x)) is (u'(x)=1).

Step2: Find the derivative of (v(x)) using the chain rule

Let (y = u^{3}) where (u=\cos(x)). By the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). (\frac{dy}{du} = 3u^{2}) and (\frac{du}{dx}=-\sin(x)). So (v'(x)=3\cos^{2}(x)(-\sin(x))=- 3\cos^{2}(x)\sin(x))

Step3: Substitute into the product rule formula

(h'(x)=1\times\cos^{3}(x)+x\times(-3\cos^{2}(x)\sin(x))=\cos^{3}(x)-3x\cos^{2}(x)\sin(x)=\cos^{2}(x)(\cos(x)-3x\sin(x)))

Answer:

B. (\cos^{2}(x)(\cos(x)-3x\sin(x)))