Question

let ( h(x)=x cos ^{3}(x) ).
find ( h^{prime}(x) ).
choose 1 answer:
(a) ( cos ^{2}(x)(cos (x)+3 x) )
(b) ( cos ^{2}(x)(cos (x)-3 x sin (x)) )
(c) ( cos ^{3}(x)-x sin ^{3}(x) )
(d) ( -3 cos ^{2}(x) sin (x) )

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