How do you use the chain rule to differentiate #root9(-cosx)#?

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Answer 1 · 2016-08-03T19:55:10

#= (sin(x))/(9(-cos(x))^(8/9))#

We have #y = (-cos(x))^(1/9)#

Let #u = -cos(x) implies (du)/(dx) = sin(x)#

#y = u^(1/9)# so using power rule:

#(dy)/(du) = 1/9(u)^(-8/9)#

#(dy)/(dx) = (dy)/(du)(du)/(dx)#

#(dy)/(dx) = 1/9(-cos(x))^(-8/9)*sin(x)#

#= (sin(x))/(9(-cos(x))^(8/9))#