How do you find the 1st and 2nd derivative of x^2e^x?

Nov 14, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(x + 2\right) {e}^{x}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left({x}^{2} + 4 x + 2\right) {e}^{x}$

Explanation:

By the product rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \times {e}^{x} + {e}^{x} \times {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {e}^{x} + {x}^{2} {e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(x + 2\right) {e}^{x}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 \times {e}^{x} + 2 x \times {e}^{x} + 2 x \times {e}^{x} + {x}^{2} \times {e}^{x}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 {e}^{x} + 2 x {e}^{x} + 2 x {e}^{x} + {x}^{2} {e}^{x}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 {e}^{x} + 4 x {e}^{x} + {x}^{2} {e}^{x}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left({x}^{2} + 4 x + 2\right) {e}^{x}$

Hopefully this helps!