Write the following function in the form \( y=f(u) \) and \( u=g(x) \). Th \( y=\left(\frac{x^{2}}{6}+5 x-\frac{4}{x}\right)^{3} \) C. \( y=f(u)=u^{3} \) and \( u=g(x)=\frac{x^{2}}{6}+5 x-\frac{4}{x} \) D. \( y=f(u)=u^{3} \) and \( u=g(x)=\left(\frac{x^{2}}{6}+5 x-\frac{4}{x}\right)^{3} \) Find \( \frac{d y}{d x} \) as a function of \( x \). \( \frac{d y}{d x}=\square \)

We are given the function \[ y=\left(\frac{x^{2}}{6}+5 x-\frac{4}{x}\right)^{3}. \] Define \[ u=g(x)=\frac{x^{2}}{6}+5x-\frac{4}{x} \] and \[ y=f(u)=u^{3}. \] To find \(\frac{dy}{dx}\), apply the chain rule: \[ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}. \] Step 1: Differentiate \(y=f(u)=u^3\) with respect to \(u\): \[ \frac{dy}{du}=3u^2. \] Step 2: Differentiate \(u=g(x)=\frac{x^2}{6}+5x-\frac{4}{x}\) with respect to \(x\): Differentiate each term separately: 1. \(\frac{d}{dx}\left(\frac{x^2}{6}\right)=\frac{1}{6}\cdot 2x=\frac{x}{3}\), 2. \(\frac{d}{dx}(5x)=5\), 3. \(\frac{d}{dx}\left(-\frac{4}{x}\right)=-4\cdot\frac{d}{dx}\left(\frac{1}{x}\right)=-4\left(-\frac{1}{x^2}\right)=\frac{4}{x^2}\). Thus, \[ \frac{du}{dx}=\frac{x}{3}+5+\frac{4}{x^2}. \] Step 3: Combine the results using the chain rule: \[ \frac{dy}{dx}=3u^2\left(\frac{x}{3}+5+\frac{4}{x^2}\right). \] Substitute back \(u=\frac{x^{2}}{6}+5x-\frac{4}{x}\): \[ \frac{dy}{dx}=3\left(\frac{x^{2}}{6}+5x-\frac{4}{x}\right)^2\left(\frac{x}{3}+5+\frac{4}{x^2}\right). \] Thus, the derivative \(\frac{dy}{dx}\) as a function of \(x\) is \[ \boxed{\frac{dy}{dx}=3\left(\frac{x^{2}}{6}+5x-\frac{4}{x}\right)^2\left(\frac{x}{3}+5+\frac{4}{x^2}\right)}. \]