What is the second derivative of $x^{2} sin x$ $x^2sinx$ ?

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What is the second derivative of $x^{2} sin x$ $x^2sinx$ ?

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Answer 1 · 2018-04-25T07:25:08

$\frac{d^{2} y}{{d x}^{2}} = 2 sin x + 4 x cos x - x^{2} sin x$ $(d^2y)/(dx^2)=2sinx+4xcosx-x^2sinx$

Explanation:

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

$\frac{d y}{d x} = \frac{d}{d x} [x^{2}] sin x + x^{2} \frac{d}{d x} [sin x] = 2 x sin x + x^{2} cos x$ $dy/dx=d/dx[x^2]sinx+x^2d/dx[sinx]=2xsinx+x^2cosx$

$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} [2 x] sin x + 2 x \frac{d}{d x} [sin x] + \frac{d}{d x} [x^{2}] cos x + x^{2} \frac{d}{d x} [cos x] = 2 sin x + 2 x cos x + 2 x cos x - x^{2} sin x = 2 sin x + 4 x cos x - x^{2} sin x$ $(d^2y)/(dx^2)=d/dx[2x]sinx+2xd/dx[sinx]+d/dx[x^2]cosx+x^2d/dx[cosx]=2sinx+2xcosx+2xcosx-x^2sinx=2sinx+4xcosx-x^2sinx$

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What is the second derivative of $x^{2} sin x$ $x^2sinx$ ?

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What is the second derivative of $x^{2} sin x$ $x^2sinx$ ?