How do you find the derivative of #y = 6 cos(x^3 + 3)# using the chain rule?

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Answer 1 · 2015-11-11T16:54:47

#(dy)/(dx) = -18x^2sin(x^3+3)#

#"Let " u=x^3+3 -> (du)/(dx)=3x^2#

#"Let " v= cos(u) -> (dv)/(du) = -sin(u)#

#"Let " y= 6v -> (dy)/(dv) = 6#

Target is #(dy)/(dx)#

By cancelling out #(dy)/(dx) = (dy)/(dv) times (dv)/(du) times (du)/(dx)#

#(dy)/(dx) = (6) times {-sin(u)} times (3x^2)#

#(dy)/(dx) = (6) times (-1) times (3) times {sin(u)} times {x^2}#

#(dy)/(dx) = -18x^2sin(x^3+3)#