# What is the derivative of #xsinx#?

##### 1 Answer

Sep 15, 2017

#### Answer:

# 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

# { ("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