How do you find the derivative of #g(x)= 3tan4xsin2xcos2x#?

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Answer 1 · 2015-05-14T09:03:44

First just remember

#sin(a)*cos(a)=1/2sin(2a)#

#3/2tan(4x)sin(4x)#

#=3/2sin^2(4x)/cos(4x)#

Remember #sin^2(a) = 1-cos^2(a)#

#3/(2cos(4x))-3/2cos(4x)#

#(3/2cos^-1(4x))' = 3/2(cos^-1(4x))' = 6sin(4x)*cos^-2(4x)#

#(u^n)'=n*u'*u^(n-1)#

#(3/2cos(4x))'=3/2(cos(4x))' = -6sin(4x)#

So the result is

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

#6sin(4x)(cos^-2(4x)+1)#