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

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Answer 1 · 2017-03-14T17:19:43

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

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

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