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 $(dy)/(dx) = (dy)/(dv) times (dv)/(du) times (du)/(dx)$

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

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

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

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