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Math Help - differentiation using quotient rule

  1. #1
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    differentiation using quotient rule

    Hello, I am stumped on how to differentiate this function:

    (1+x) / x^4

    My attempt is:
    1(x^4) - (1+x)4x^3
    = (x^4 - (4x^3+4x^4)) / x^8
    = ((-4x^3)-4) / x*

    at this point I am not sure if I am headed in the right direction.

    Thanks for any help.
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  2. #2
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    Re: differentiation using quotient rule

    Quote Originally Posted by fran1942 View Post
    Hello, I am stumped on how to differentiate this function:

    (1+x) / x^4

    My attempt is:
    1(x^4) - (1+x)4x^3
    = (x^4 - (4x^3+4x^4)) / x^8
    = ((-4x^3)-4) / x*

    at this point I am not sure if I am headed in the right direction.

    Thanks for any help.
    \frac{1+x}{x^4} = \frac{1}{x^4} + \frac{x}{x^4} = x^{-4} + x^{-3}

    now use the power rule
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  3. #3
    Super Member Quacky's Avatar
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    Re: differentiation using quotient rule

    Please make an effort to learn latex - your post is extremely difficult to follow.

    We get f'(x)=\frac{(x^4) - (1+x)4x^3}{x^8}

    =\frac{(x^4) - 4x^3-4x^4}{x^8}

    =\frac{-4x^3-3x^4}{x^8}

    Now, as far as I can see, once you divide through a common factor, you're done.
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  4. #4
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    Re: differentiation using quotient rule

    thanks kindly, but I just cant see how I end up with a denominator of x^5 as per the textbook answer.
    I thought the common factor and final answer denominator would be x^4 ?
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  5. #5
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    Re: differentiation using quotient rule

    Quote Originally Posted by fran1942 View Post
    thanks kindly, but I just cant see how I end up with a denominator of x^5 as per the textbook answer.
    I thought the common factor and final answer denominator would be x^4 ?
    There is a common factor of x^3 in the numerator. When you cancel, you are left with x^5 in the denominator.
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