What are the first three derivatives of $(xcos(x)-sin(x))/(x^2)$ ?

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Answer 1 · 2015-03-12T19:22:34

The answer is:

$y''=(-x^3cosx+3x^4sinx+6xcosx-6sinx)/x^4$ .

This is why:

$y'=(((cosx+x*(-sinx)-cosx)x^2-(xcosx-sinx)*2x))/x^4=$

$=(-x^3sinx-2x^2cosx+2xsinx)/x^4=$

$=(-x^2sinx-2xcosx+2sinx)/x^3$

$y''=((-2xsinx-x^2cosx-2cosx-2x(-sinx)+2cosx)x^3-(-x^2sinx-2xcosx+2sinx)*3x^2)/x^6=$

$=((-x^2cosx)x^3+3x^4sinx+6x^3cosx-6x^2sinx)/x^6=$

$=(-x^3cosx+3x^4sinx+6xcosx-6sinx)/x^4$ .

What are the first three derivatives of (xcos(x)-sin(x))/(x^2)(xcos(x)-sin(x))/(x^2)?