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#