The problem asks to find the first and second derivatives of the function $y = \frac{x^4}{2} - \frac{3x^2}{2} - x$.

1. Problem Description

The problem asks to find the first and second derivatives of the function $y = \frac{x^4}{2} - \frac{3x^2}{2} - x$.

2. Solution Steps

We are given the function $y = \frac{x^4}{2} - \frac{3x^2}{2} - x$.

First, we find the first derivative, denoted as $\frac{dy}{dx}$ or $y'$. We use the power rule for differentiation, which states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$.

$\frac{dy}{dx} = \frac{d}{dx}(\frac{x^4}{2}) - \frac{d}{dx}(\frac{3x^2}{2}) - \frac{d}{dx}(x)$

$\frac{dy}{dx} = \frac{1}{2}\frac{d}{dx}(x^4) - \frac{3}{2}\frac{d}{dx}(x^2) - \frac{d}{dx}(x)$

$\frac{dy}{dx} = \frac{1}{2}(4x^3) - \frac{3}{2}(2x) - 1$

$\frac{dy}{dx} = 2x^3 - 3x - 1$

Now, we find the second derivative, denoted as $\frac{d^2y}{dx^2}$ or $y''$. We differentiate the first derivative with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x^3 - 3x - 1)$

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x) - \frac{d}{dx}(1)$

$\frac{d^2y}{dx^2} = 2\frac{d}{dx}(x^3) - 3\frac{d}{dx}(x) - 0$

$\frac{d^2y}{dx^2} = 2(3x^2) - 3(1)$

$\frac{d^2y}{dx^2} = 6x^2 - 3$

3. Final Answer

The first derivative is $\frac{dy}{dx} = 2x^3 - 3x - 1$.

The second derivative is $\frac{d^2y}{dx^2} = 6x^2 - 3$.