How can I find the derivative of this?

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Answer 1 · 2017-03-18T16:15:42

$(d^2y)/(dx^2)=6xcosx-6x^2sinx-x^3cosx$

As $y=x^3cosx$

$(dy)/(dx)=3x^2cosx+x^3(-sinx)$

= $3x^2cosx-x^3sinx$

and $(d^2y)/(dx^2)=d/(dx)((dy)/(dx))$

= $d/(dx)(3x^2cosx-x^3sinx)$

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

= $6xcosx-3x^2sinx-3x^2sinx-x^3cosx$

= $6xcosx-6x^2sinx-x^3cosx$