Using the product rule, differentiate \(y=(3x-2x^2)(5+4x)\).

To differentiate \(y=(3x-2x^2)(5+4x)\) using the product rule, we need to differentiate each term separately and then add them together.

Let's denote \(u = 3x - 2x^2\) and \(v = 5 + 4x\).

Now, we can differentiate \(u\) and \(v\) with respect to \(x\):

\(\frac{du}{dx} = 3 - 4x\)

\(\frac{dv}{dx} = 4\)

Now, we can apply the product rule:

\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)

\(\frac{dy}{dx} = (3x - 2x^2)(4) + (5 + 4x)(3 - 4x)\)

\(\frac{dy}{dx} = 12x - 8x^2 + 15 - 20x + 12x - 16x^2\)

\(\frac{dy}{dx} = -24x^2 + 24x + 15\)

Therefore, the derivative of \(y=(3x-2x^2)(5+4x)\) is \(\frac{dy}{dx} = -24x^2 + 24x + 15\).

Using the product rule, we have:
$$\frac{d}{dx}[(3x-2x^2)(5+4x)] = (3x-2x^2)\frac{d}{dx}[5+4x] + (5+4x)\frac{d}{dx}[3x-2x^2]$$
$$=(3x-2x^2)(4) + (5+4x)(3-4x)$$
$$=12x-8x^2 + 15-20x -3x^2-16x^2$$
$$=-3x^2-24x+15$$