Question

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

Answer 1

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`.