Question

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

Answer 1

`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`