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Math Help - Intro to Derivatives - Quotient Rule

  1. #1
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    Intro to Derivatives - Quotient Rule



    There is no answer in my textbook for this question.
    Please show a step by step solution so I could follow through...

    Here's what I've done so far:

    y' = \frac {5(x + 2) - 1(5x + 2)}{(x+2)^2}

    y' = \frac {5x + 10 - 5x - 2}{(x+2)^2}

    y' = \frac {8}{(x+2)^2}

    Now what?
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  2. #2
    o_O
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    You know that derivatives essentially represent the slopes of a function. So, if you were to have tangents with a negative slope, f'(x) must be negative.

    If f(x) does not have any negative tangents, that must mean f'(x) is never less than 0. Does this happen in your expression for f'(x)?
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  3. #3
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    Quote Originally Posted by o_O View Post
    You know that derivatives essentially represent the slopes of a function. So, if you were to have tangents with a negative slope, f'(x) must be negative.

    If f(x) does not have any negative tangents, that must mean f'(x) is never less than 0. Does this happen in your expression for f'(x)?
    Yes, it doesn't have any negative tangents because the denominator will always be positive and the numerator is a constant positive value?

    OT: When you're finding the x values to substitute in the original equation, do you solve for the numerator or denominator?

    For example;

    Original Equation: f(x) = ax^2 + bx + c

    f'(x) = \frac {(x+4)(x-5)}{(x-2)^2}
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  4. #4
    o_O
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    .. I'm assuming you took the derivative right
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