What is the second derivative of x^2sinx?

Apr 25, 2018

Answer:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 \sin x + 4 x \cos x - {x}^{2} \sin x$

Explanation:

We have $y = {x}^{2} \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] \sin x + {x}^{2} \frac{d}{\mathrm{dx}} \left[\sin x\right] = 2 x \sin x + {x}^{2} \cos x$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left[2 x\right] \sin x + 2 x \frac{d}{\mathrm{dx}} \left[\sin x\right] + \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] \cos x + {x}^{2} \frac{d}{\mathrm{dx}} \left[\cos x\right] = 2 \sin x + 2 x \cos x + 2 x \cos x - {x}^{2} \sin x = 2 \sin x + 4 x \cos x - {x}^{2} \sin x$