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

May 21, 2026Mathematics

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Answer

\mathbf{6x^2+2x-17}

Graph of y = (x² + 3x − 1)(2x − 5)

4 steps

Identify $u$ and $v$ from the product $y = uv$ .
Let $u = x^2+3x-1$ and $v = 2x-5$ .
Differentiate $u$ and $v$ with respect to $x$ .
$\frac{du}{dx} = \frac{d}{dx}(x^2+3x-1) = 2x+3$ $\frac{dv}{dx} = \frac{d}{dx}(2x-5) = 2$
Apply the product rule for differentiation, $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ .
$\frac{dy}{dx} = (x^2+3x-1)(2) + (2x-5)(2x+3)$
Expand and simplify the expression.
$\frac{dy}{dx} = 2(x^2+3x-1) + (2x-5)(2x+3)$ $\frac{dy}{dx} = 2x^2+6x-2 + (4x^2+6x-10x-15)$ $\frac{dy}{dx} = 2x^2+6x-2 + 4x^2-4x-15$ $\frac{dy}{dx} = (2x^2+4x^2) + (6x-4x) + (-2-15)$ $\frac{dy}{dx} = 6x^2+2x-17$
The final answer is $\mathbf{6x^2+2x-17}$ .

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