Given y = (x2+3x−1)(2x−5), we can use the product rule for differentiation.

Given $y = (x^2+3x-1)(2x-5)$, we can use the product rule for differentiation.
Let $u = x^2+3x-1$ and $v = 2x-5$.
Then, find the derivatives of $u$ and $v$ with respect to $x$:

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

The product rule states that if $y = uv$, then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$.
Substitute the expressions for $u, v, \frac{du}{dx},$ and $\frac{dv}{dx}$ into the 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:

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

The final answer is $6x^2+2x-17$.