Differentiate the function y = (x2+3x−1)(2x−5) using the product rule.

Let $u = x^2+3x-1$ and $v = 2x-5$.
Differentiate $u$ with respect to $x$:

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

Differentiate $v$ with respect to $x$:

$$\frac{dv}{dx} = \frac{d}{dx}(2x-5) = 2$$

Substitute the expressions for $u, v, \frac{du}{dx},$ and $\frac{dv}{dx}$ into the product rule formula:

$$\frac{dy}{dx} = (x^2+3x-1)(2) + (2x-5)(2x+3)$$

$$\frac{dy}{dx} = 2(x^2+3x-1) + (2x-5)(2x+3)$$ $$\frac{dy}{dx} = 2x^2+6x-2 + (4x^2+6x-10x-15)$$ $$\frac{dy}{dx} = 2x^2+6x-2 + 4x^2-4x-15$$

Combine like terms:

$$\frac{dy}{dx} = (2x^2+4x^2) + (6x-4x) + (-2-15)$$ $$\frac{dy}{dx} = 6x^2+2x-17$$