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

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

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Answer 1 · 2015-11-11T16:54:47

$\frac{d y}{d x} = - 18 x^{2} sin (x^{3} + 3)$ $(dy)/(dx) = -18x^2sin(x^3+3)$

Explanation:

$Let u = x^{3} + 3 \to \frac{d u}{d x} = 3 x^{2}$ $"Let " u=x^3+3 -> (du)/(dx)=3x^2$

$Let v = cos (u) \to \frac{d v}{d u} = - sin (u)$ $"Let " v= cos(u) -> (dv)/(du) = -sin(u)$

$Let y = 6 v \to \frac{d y}{d v} = 6$ $"Let " y= 6v -> (dy)/(dv) = 6$

Target is $\frac{d y}{d x}$ $(dy)/(dx)$

By cancelling out $\frac{d y}{d x} = \frac{d y}{d v} \times \frac{d v}{d u} \times \frac{d u}{d x}$ $(dy)/(dx) = (dy)/(dv) times (dv)/(du) times (du)/(dx)$

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

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

$\frac{d y}{d x} = - 18 x^{2} sin (x^{3} + 3)$ $(dy)/(dx) = -18x^2sin(x^3+3)$

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

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the derivative of y=6cos(x3+3)y = 6 cos(x^3 + 3) using the chain rule?

1 Answer

Explanation:

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Impact of this question

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