`f(x) = (5x^2 + 8)(x^2 - 4x - 6)` Use the Product Rule or the Quotient Rule to find the derivative of the function.

Textbook Question

Chapter 2, Review - Problem 29 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the derivative of the given function and since the function is a product of two polynomials, then you must use the product rule, such that:

`f'(x) = (5x^2 + 8)'(x^2 - 4x - 6) + (5x^2 + 8)(x^2 - 4x - 6)'`

`f'(x) = (10x + 0)(x^2 - 4x - 6) + (5x^2 + 8)(2x - 4 - 0)`

`f'(x) = 10x(x^2 - 4x - 6) + (2x - 4)(5x^2 + 8)`

`f'(x) = 10x^3 - 40x^2 - 60x + 10x^3 + 16x - 20x^2 - 32`

Combining like terms yields:

`f'(x) = 20x^3 - 60x^2 - 44x - 32`

Hence, evaluating the derivative of the function, using the product rule, yields `f'(x) = 20x^3 - 60x^2 - 44x - 32.`

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loves2learn | (Level 3) Salutatorian

Posted on

Using a Quotient rule:

Given,

Then,

Therefore, by bringing `x^2-4x-6` up using a negative exponent, you get

`y=(5x^2+8)/(x^2-4x-6)^-1 `

`y'=((10x)(x^2-4x-6)^-1-(5x^2+8)(-1)(x^2-4x-6)^-2(2x-4))/(x^2-4x-6)^-2 `

Simplify all of this,

`y'=((10x)(x^2-4x-6)^-1+(2x-4)(5x^2)(x^2-4x-6)^-2)/(x^2-4x-6)^-2 `

`y'=20x^3-60x^2-44x-32 `

Note that you should get the same answer if you used a product rulet. If you don't, then check your answer because you must have made a mistake along the way.

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