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 # 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.