`y = (x^2 - 3x + 2)(x^3 + 1), c=2` Find f'(x) and f'(c).

Textbook Question

Chapter 2, 2.3 - Problem 14 - 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 first y', using the product rule, such that:

`y' = (x^2 - 3x + 2)'(x^3 + 1) + (x^2 - 3x + 2)(x^3 + 1)'`

`y' = (2x - 3)(x^3 + 1) + (x^2 - 3x + 2)(3x^2)`

`y' = 2x^4 + 2x - 3x^3 - 3 + 3x^4 - 9x^3 + 6x^2`

Combining like terms yields:

`y' = 5x^4 - 12x^3 + 6x^2 + 2x - 3`

You may evaluate now f'(c) at c = 2, replacing 2 for x in equation of f'(x):

`y' = 5*2^4 - 12*2^3 + 6*2^2 + 2*2 - 3`

`y' = 80 - 96 + 24 + 4 - 3`

`y' = 105 - 96 => y' = 9`

Hence, evaluating the derivative of the function yields `y' = 5x^4 - 12x^3 + 6x^2 + 2x - 3` and evaluating the value of derivative at c= 2, yields y' = 9.

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