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Math Help - VERY VERY VERY simple derivative question

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    VERY VERY VERY simple derivative question

    I am very confused on this although it should be very basic. The derivative of (4-x)^-1 is (4-x)^-2, since you use the chain rule. Then why is the derivative of (4-x)^-2 not 2(4-x)^-3, but -2(4-x)^-3? Shouldn't it be positive since you multiply the -3 to the front and then multiply by the derivative of the inside function (4-x) so you multiply by -1 making it positive? I'm so confused?
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  2. #2
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    Quote Originally Posted by virtuoso735 View Post
    I am very confused on this although it should be very basic. The derivative of (4-x)^-1 is (4-x)^-2, since you use the chain rule. Then why is the derivative of (4-x)^-2 not 2(4-x)^-3, but -2(4-x)^-3? Shouldn't it be positive since you multiply the -3 to the front and then multiply by the derivative of the inside function (4-x) so you multiply by -1 making it positive? I'm so confused?
    The chain rule says that for h(x) = f(g(x)) then h'(x) = g'(x)f'(g(x))

    y = (4-x)^{-2}

    f(x) = (4-x)^{-2}
    g(x) = (4-x)

    therefore we get y' = -1 \times -2(4-x)^{-3} = 2(4-x)^{-3}

    It would therefore seem your answer is correct
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  3. #3
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    Okay, so the first derivative of (4-x)^-1 is (4-x)^-2.
    The second derivative should be 2(4-x)^-3, right?
    And the third derivative should be 6(4-x)^-4?

    But the wolfram online derivator gives me -2(4-x)^-3 for the second derivative?!
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  4. #4
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    I've just tried the Wolfram derivative calculator and for the second derivative of (4-x)^{-2} it gives:

    -2 \, (x-4)^{-3} which is the same as 2 \, (4-x)^{-3} hence no actual contradiction if you look more carefully inside the brackets. The negative sign is just because it swaps the subtraction order within the brackets. Computational algebra for you!
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  5. #5
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    Woops, didn't realize that they switched the x and 4. Makes sense now. Thanks.
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