Find the first and second derivatives of the function $y = \frac{1+3x}{3x}(3-x)$.

1. Problem Description

Find the first and second derivatives of the function $y = \frac{1+3x}{3x}(3-x)$.

2. Solution Steps

First, simplify the expression for $y$:

$y = \frac{1+3x}{3x}(3-x) = \frac{(1+3x)(3-x)}{3x} = \frac{3 - x + 9x - 3x^2}{3x} = \frac{3 + 8x - 3x^2}{3x} = \frac{3}{3x} + \frac{8x}{3x} - \frac{3x^2}{3x} = \frac{1}{x} + \frac{8}{3} - x$

Now, find the first derivative $y'$:

$y' = \frac{d}{dx}(\frac{1}{x} + \frac{8}{3} - x) = \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(\frac{8}{3}) - \frac{d}{dx}(x)$

$\frac{d}{dx}(x^{-1}) = -1x^{-2} = -\frac{1}{x^2}$

$\frac{d}{dx}(\frac{8}{3}) = 0$

$\frac{d}{dx}(x) = 1$

Therefore,

$y' = -\frac{1}{x^2} + 0 - 1 = -\frac{1}{x^2} - 1$

Next, find the second derivative $y''$:

$y'' = \frac{d}{dx}(-\frac{1}{x^2} - 1) = \frac{d}{dx}(-x^{-2}) - \frac{d}{dx}(1)$

$\frac{d}{dx}(-x^{-2}) = -(-2)x^{-3} = 2x^{-3} = \frac{2}{x^3}$

$\frac{d}{dx}(1) = 0$

Therefore,

$y'' = \frac{2}{x^3} - 0 = \frac{2}{x^3}$

3. Final Answer

The first derivative is $y' = -\frac{1}{x^2} - 1$.

The second derivative is $y'' = \frac{2}{x^3}$.