What is the derivative of xsinx?

1 Answer
Sep 15, 2017

dy/dx = xcosx+sinx

Explanation:

We have:

y = xsin x

Which is the product of two functions, and so we apply the Product Rule for Differentiation:

d/dx(uv)=u(dv)/dx+(du)/dxv , or, (uv)' = (du)v + u(dv)

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

So with y = xsinx ;

{ ("Let", u = x, => (du)/dx = 1), ("And" ,v = sinx, => (dv)/dx = cosx ) :}

Then:

d/dx(uv)=u(dv)/dx + (du)/dxv

Gives us:

d/dx( xsinx) = (x)(cosx)+(1)(sinx)
:. dy/dx = xcosx+sinx

If you are new to Calculus then explicitly substituting u and v can be quite helpful, but with practice these steps can be omitted, and the product rule can be applied as we write out the solution.