`y = root(5)(3x^3 + 4x), (2,2)` Find and evaluate the derivative of the function at the given point.

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Chapter 2, 2.4 - Problem 66 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the derivative of the function, using the chain rule, such that:

`y' = ((3x^3 + 4x)^(1/5))' => y' = ((1/5)(3x^3 + 4x)^(1/5 - 1))(3x^3 + 4x)'`

`y' = ((1/5)(3x^3 + 4x)^(-4/5))(9x^2 + 4)`

`y' = (9x^2 + 4)/(5 root(5)((3x^3 + 4x)^4))`

Now, you need to evaluate the value of derivative at x = 2:

`y' = (9*2^2 + 4)/(5 root(5)((3*2^3 + 4*2)^4))`

`y' = (36 + 4)/(5 root(5)((24 + 8)^4))`

`y' = 40/(5*root(5) (32^4)) => y' = 40/(5*root(5) ((2^5)^4))`

`y' = 40/(5*root(5) (2^20)) => y' = 40/(5*(2^(20/5))) => y' = 40/(5*2^4)`

`y' = 40/80`

Simplifying by 40 yields:

`y' = 1/2`

Hence, evaluating the derivative of the function yields `y' = (9x^2 + 4)/(5 root(5)((3x^3 + 4x)^4))` and evaluating the value of derivative at x = 2, yields `y' = 1/2.`

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