How do you differentiate #f(x)= xsin3x# using the product rule?

1 Answer
Feb 2, 2016

Answer:

#sin3x+3xcos3x#

Explanation:

The product rule states that:
#d/dxu(x)v(x)=u'(x)v(x)+u(x)v'(x)#

In our example, #u(x)# (the first function) is simply #x#, while #v(x)# (the other function) is #sin3x#:
#x*sin3x#
#color(white)(AAA)#
#u(x)color(white)()v(x)#

Let's compute the derivatives of #x# and #sin3x# here:
#x'=1#
#(sin3x)'=3cos3x->#(by the chain rule)

Now we plug everything into the chain rule equation:
#d/dxxsin3x=(x)'(sin3x)+(x)(sin3x)'#
#d/dxxsin3x=(1)(sin3x)+(x)(3cos3x)#
#d/dxxsin3x=sin3x+3xcos3x#