How do you differentiate #f(x)=1/(x^3-1)x^2cosx# using the product rule?

Question

1109 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T11:28:53

#f'(x)=-(x^4+2x)/(x^3-1)^2*cos(x)-x^2/(x^3-1)*sin(x)#

#(uv)'=u'v+uv'# we get with
#u=x^2/(x^3-1)#

#u'=(2x(x^3-1)-x^2*3x^2)/(x^3-1)^2#
simplifying we get

#u'=-(x^4+2x)/(x^3-1)^2#
#v=cos(x)#
#v'=-sin(x)#

putting thins together we get

#f'(x)=-(x^4+2x)/(x^3-1)^2cos(x)-x^2/(x^3-1)sin(x)#

How do you differentiate #f(x)=1/(x^3-1)*x^2*cosx# using the product rule?