How do you find the derivative $y=xsinx + cosx$ ?

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Answer 1 · 2015-02-26T01:19:29

$y'=xcosx$

Use the product rule for $d((xsinx))/(dx)$ :

$(uv)'=u(dv)/(dx)+v(du)/(dx)$

Where $u=x$ and $v=sinx$

So $(d(xsinx))/(dx)=xcosx+sinx(dx)/(dx)=xcosx+sinx$

Now $(d(cosx))/(dx)=-sinx$

So combining the 2 we get:

$(d(xsinx+cosx))/(dx)=xcosx+sinx-sinx=xcosx$

How do you find the derivative y=xsinx + cosxy=xsinx + cosx?