`y = 26 - (sec(4x))^3, (0,25)` Find and evaluate the derivative of the function at the given point. Use a graphing utility to verify your result.

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Chapter 2, 2.4 - Problem 71 - 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' = (26 - (sec(4x))^3)' => y' = -3(sec(4x))^2* (sec(4x))*(tan(4x))*(4x)'`

`y' = -12(sec(4x))^3*(tan(4x))`

Now, you need to evaluate the value of derivative at x = 0:

`y' = -12(sec(0))^3*(tan(0))`

`y' = -12*(1/(cos^3 0))*0`

`y' = 0`

Hence, evaluating the derivative of the function yields `y' = -12(sec(4x))^3*(tan(4x))` and evaluating the value of derivative at x = 0, yields `y' =0` .

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