How do you find the derivative of #x^2 sinx#?

1 Answer
May 22, 2016

#d/dx(x^2sinx)=2xsinx+x^2cosx#

Explanation:

The product rule states that:
#d/dx(uv)=u'v+uv'#
Where #u# and #v# are functions of #x#.

In #x^2sinx#, we have two functions: #x^2# and #sinx#. Since they are being multiplied together, we'll need to use the product rule to find the derivative.

Let #u=x^2# and #v=sinx#:
#u=x^2->u'=2x#
#v=sinx->v'=cosx#

Making necessary substitutions in the product rule, we have:
#(2x)(sinx)+(x^2)(cosx)#
#=2xsinx+x^2cosx#

We can't really simplify this further, so we'll leave it as our final answer.