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Math Help - Finding the derivatives of functions

  1. #1
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    Finding the derivatives of functions

    I have 2 problems I was confused with.
    1. f(x)=(2x2-3x+1)/x
    2. f(x)=x4/5

    This is what I did for the first one:
    I moved the x up, which turns into 2x2-3x+1+x-1
    Then I multipled the coefficient by the exponents, and subtracted 1 from the exponent: 4x-3+1+x-2
    But I was wondering, what do you do for the 1 here?

    And for the second one:
    Same thing, multipled exponent into coefficient: (4/5)x-1/5
    Then 1/(4/5x5)
    Then I got lost here... the answer in the book is: 4/(5x1/5)
    It looks like the denominator just had the exponent switched into the positive, is that the right way to move it though?
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  2. #2
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    Re: Finding the derivatives of functions

    your algebra needs work ...


    \frac{2x^2 - 3x + 1}{x} = \frac{2x^2}{x} - \frac{3x}{x} + \frac{1}{x} = 2x - 3 + x^{-1} , now take the derivative.


    \frac{4}{5} x^{-\frac{1}{5}} = \frac{4}{5x^{\frac{1}{5}}} , only the x has a negative exponent
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    Re: Finding the derivatives of functions

    Quote Originally Posted by skeeter View Post
    your algebra needs work ...


    \frac{2x^2 - 3x + 1}{x} = \frac{2x^2}{x} - \frac{3x}{x} + \frac{1}{x} = 2x - 3 + x^{-1} , now take the derivative.


    \frac{4}{5} x^{-\frac{1}{5}} = \frac{4}{5x^{\frac{1}{5}}} , only the x has a negative exponent
    Oh thanks!
    So then the derivative would be 2-x-2, which then becomes 2-(1/x2) right?

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  4. #4
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    Re: Finding the derivatives of functions

    Quote Originally Posted by Chaim View Post
    So then the derivative would be 2-x-2, which then becomes 2-(1/x2) right?
    correct
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    Re: Finding the derivatives of functions

    Quote Originally Posted by skeeter View Post
    correct
    Thanks!

    Now I think I'm starting to get how to find derivatives of a function.
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