How do you differentiate #f(x)=sin^2(lnx)xcos^2(x^2)# using the chain rule?

1 Answer
May 15, 2017

Answer:

#=-4x^2sin^2(lnx)sin(x^2)cos(x^2)+sin^2(lnx)cos^2(x^2)+xsin(2lnx)cos^2(x^2)

Explanation:

#=[sin^2(lnx)][d/dx(xcos^2(x^2))]+[d/dx(sin^2(lnx))][xcos^2(x^2)]#
#=[sin^2(lnx)][(x)(d/dx(cos^2(x^2)))+(1)(cos^2(x^2))]+[(2sin(lnx)cos(lnx))/x][cos^2(x^2)]#
#=[sin^2(lnx)][(x)(-2sin(x^2)cos(x^2)(2x))+cos^2(x^2)]+[sin(2lnx)cos^2(x^2)]#
#=-4x^2sin^2(lnx)sin(x^2)cos(x^2)+sin^2(lnx)cos^2(x^2)+xsin(2lnx)cos^2(x^2)#

The chain rule is very messy here but this should guide you:
For #sin^2(lnx)#, differentiate the quadratic term, triginometric term, #ln# term and lastly the #x# term.

For #xsin^2(x^2)# Use Product Rule to separate #x#. Differentiate the quadratic term, then the trigonometric term then the inside #(x^2)# term.
#d/dx[cos^2(x^2)]->2cos(x^2)->-2cos(x^2)sin(x^2)->-4xsin(x^2)cos(x^2)#