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How do you differentiate #y = 6 cos(x^3 + 3)#?

1 Answer

Mar 14, 2017

#y' = -18x^2 sin(x^3 + 3)#

Explanation:

Use #(cos u)' = -sin u (u')#

Let #u = x^3 +3#, #u' = 3x^2#

If #y = 6cos(x^3 + 3)# then

#y' = 6 (-sin (x^3 + 3))(3x^2)#

Simplify: #y' = -18x^2 sin(x^3 + 3)#

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