Help: derivative and critical #'s

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The discussion focuses on finding the derivatives of two functions: f(x) = -2x - (128/(x+9)) and f(x) = (x+4)^2 (x-10)^3. The derivative of the first function is determined as f'(x) = -2 + 128/(x+9)^2, utilizing the quotient rule for differentiation. For the second function, the product rule is applied to find the derivative, with the additional application of the chain rule being noted but not prominently affecting the outcome. The goal is to identify the absolute minimum of the second function within the interval 4 ≤ x ≤ 10.

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buffgilville
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I can't seem to find the f ' (x) of the following:

1) f(x) = -2x - ((128)/(x+9))

2) f(x) = (x+4)^2 (x-10)^3
(Find the number x at which the function f(x) = (x+4)^2 (x-10)^3 takes on an absolute minimum on the interval 4 < x < 10 (note: < means greater than or equal to )
 
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1. It's just -2+128/(x+9)^2. Do you understand why?
2. Do you know how to take the derivative of a product of two functions?
 
buffgilville said:
I can't seem to find the f ' (x) of the following:

1) f(x) = -2x - ((128)/(x+9))

2) f(x) = (x+4)^2 (x-10)^3
(Find the number x at which the function f(x) = (x+4)^2 (x-10)^3 takes on an absolute minimum on the interval 4 < x < 10 (note: < means greater than or equal to )


The derivative of -2x is -2. To differentiate ((128)/(x+9) use the quotient rule.

To differentiate (x+4)^2 (x-10)^2 you will need to use the product rule.
(and the chain rule but the derivatives of both x+4 and x- 10 are 1 so it doesn't really show up.)
 

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