What is the derivative of #f(x) = xcos^3(x^2)sin(x^2)#?

1 Answer
Oct 20, 2016

#(df)/(dx)=cos^3(x^2)sin(x^2)-6x^2cos^2(x^2)sin^2(x^2)+2x^2cos^4(x^2)#

Explanation:

We can use here product formula for three terms i.e.

if #f(x)=p(x)*q(x)*r(x)#, then

#f'(x)= p'(x)*q(x)*r(x)+p(x)*q'(x)*r(x)+p(x)*q(x)*r'(x)#

Hence for given function #f(x)=xcos^3(x^2)sin(x^2)#

#(df)/(dx)=1xxcos^3(x^2)sin(x^2)+x*(3cos^2(x^2)xx(-sin(x^2))xx2x)*sin(x^2)+xcos^3(x^2)xx(cos(x^2)xx2x)#

= #cos^3(x^2)sin(x^2)-6x^2cos^2(x^2)sin^2(x^2)+2x^2cos^4(x^2)#