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What is the derivative of #sin^2(4x)#?

1 Answer

Nov 8, 2017

#(dy)/(dx)=8sin(4x)cos(4x)#

Explanation:

#y=(sin(4x))^2=(f(x))^2#

#y=(f(x))^n=>(dy)/(dx)=n*f'(x)*(f(x))^(n-1)#

So, #y=(sin(4x))^2#
#(dy)/(dx)=2*sin(4x)*d/(dx)sin(4x)#
#=2sin(4x)*4cos(4x)#
#=8sin(4x)cos(4x)#

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