How do you use the chain rule to differentiate #y=sin^2(cos(4x))#?

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Answer 1 · 2017-12-11T07:11:04

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

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

#dy/dx = 2sin(cos(4x)) * d/dx sin(cos(4x))#

Apply the chain rule and standard differential

#dy/dx= 2sin(cos(4x)) * cos(cos(4x)) * d/dx cos(4x)#

Apply the chain rule and standard differential again

#dy/dx= 2sin(cos(4x)) * cos(cos(4x)) * -sin(4x) *d/dx 4x#

Apply power rule

#dy/dx= 2sin(cos(4x)) * cos(cos(4x)) * -sin(4x) *(4)#

#= -8sin(4x)sin(cos(4x))cos(cos(4x))#