Define y and y + y.

We are given $y = 5x^3 - 8x$.
Then, we replace $x$ with $x + \Delta x$ to find $y + \Delta y$:

$$y + \Delta y = 5(x + \Delta x)^3 - 8(x + \Delta x)$$

First, expand $(x + \Delta x)^3$:

$$(x + \Delta x)^3 = x^3 + 3x^2(\Delta x) + 3x(\Delta x)^2 + (\Delta x)^3$$

Substitute this back into the equation for $y + \Delta y$:

$$y + \Delta y = 5(x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3) - 8(x + \Delta x)$$ $$y + \Delta y = 5x^3 + 15x^2\Delta x + 15x(\Delta x)^2 + 5(\Delta x)^3 - 8x - 8\Delta x$$

Subtract $y$ from $y + \Delta y$:

$$\Delta y = (5x^3 + 15x^2\Delta x + 15x(\Delta x)^2 + 5(\Delta x)^3 - 8x - 8\Delta x) - (5x^3 - 8x)$$ $$\Delta y = 5x^3 + 15x^2\Delta x + 15x(\Delta x)^2 + 5(\Delta x)^3 - 8x - 8\Delta x - 5x^3 + 8x$$

Combine like terms:

$$\Delta y = 15x^2\Delta x + 15x(\Delta x)^2 + 5(\Delta x)^3 - 8\Delta x$$

Factor out $\Delta x$ from the expression for $\Delta y$:

$$\frac{\Delta y}{\Delta x} = \frac{\Delta x (15x^2 + 15x\Delta x + 5(\Delta x)^2 - 8)}{\Delta x}$$

Cancel $\Delta x$:

$$\frac{\Delta y}{\Delta x} = 15x^2 + 15x\Delta x + 5(\Delta x)^2 - 8$$

The derivative $\frac{dy}{dx}$ is defined as the limit of $\frac{\Delta y}{\Delta x}$ as $\Delta x$ approaches 0:

$$\frac{dy}{dx} = \lim_{\Delta x \to 0} (15x^2 + 15x\Delta x + 5(\Delta x)^2 - 8)$$

As $\Delta x \to 0$, the terms $15x\Delta x$ and $5(\Delta x)^2$ become 0:

$$\frac{dy}{dx} = 15x^2 + 15x(0) + 5(0)^2 - 8$$ $$\frac{dy}{dx} = 15x^2 - 8$$

The derivative of $y = 5x^3 - 8x$ is: