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

May 21, 2026Mathematics

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Answer

6x^2+2x-17

Graph of y = (x² + 3x − 1)(2x − 5)

Key points of y = (x² + 3x − 1)(2x − 5)
Point	x	y
x-intercept	−3.3	0
Maximum	−1.86	27.21
y-intercept	0	5
x-intercept	0.3	0
Minimum	1.52	−11.51
x-intercept	2.5	0

Step-by-step solution

5 steps

To find $\frac{dy}{dx}$ for $y = (x^2+3x-1)(2x-5)$ , we will use the product rule for differentiation, which states that if $y = uv$ , then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ .

Identify $u$ and $v$ .
Let $u = x^2+3x-1$
Let $v = 2x-5$
Differentiate $u$ with respect to $x$ to find $\frac{du}{dx}$ .
$\frac{du}{dx} = \frac{d}{dx}(x^2+3x-1) = 2x+3$
Differentiate $v$ with respect to $x$ to find $\frac{dv}{dx}$ .
$\frac{dv}{dx} = \frac{d}{dx}(2x-5) = 2$
Apply the product rule formula $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ .
$\frac{dy}{dx} = (x^2+3x-1)(2) + (2x-5)(2x+3)$
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 derivative is:
$6x^2+2x-17$

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

Answer

Step-by-step solution

Identify uuu and vvv.

Differentiate uuu with respect to xxx to find dudx\frac{du}{dx}dxdu​.

Differentiate vvv with respect to xxx to find dvdx\frac{dv}{dx}dxdv​.

Apply the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}dxdy​=udxdv​+vdxdu​.

Expand and simplify the expression.

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More Mathematics Questions

Identify $u$ and $v$ .

Differentiate $u$ with respect to $x$ to find $\frac{du}{dx}$ .

Differentiate $v$ with respect to $x$ to find $\frac{dv}{dx}$ .

Apply the product rule formula $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ .