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Math Help - Simplification of a derivative

  1. #1
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    Simplification of a derivative

    I'm in the process of a derivative and I'm having trouble simplifying. Would someone mind helping me out with where to start.

    Here is the original problem I was told to differentiate:

    \theta = \frac{1-a\sqrt[3]{t^2}}{1+a\sqrt{t^3}}

    I decided to use the quotient rule and came up with this as the first step:

    \frac{1-at^{\frac{2}{3}}}{ 1+at^{\frac{3}{2}}}

    and then this:

    \frac{(1+at^{\frac{3}{2}})(\frac{-2}{3}at^{\frac{-1}{3}}) - (1-at^{\frac{2}{3}})(\frac{3}{2}at^{\frac{1}{2}})}{(1  +at^{\frac{3}{2}})^2}

    and here is what I multiplied it all out to:

    \frac{\frac{-2}{3}at^\frac{-1}{3} + \frac{-2}{3}a^2t^{\frac{7}{6}} - \frac{3}{2}at^{\frac{1}{2}} - \frac{3}{2}a^2t^{\frac{7}{6}}}{(1+at^{\frac{3}{2}}  )^2}

    Now, its been a while and my algebra is rusty so I'm stuck here. What is the first thing you would do to go about simplifying? Thanks in advance.
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  2. #2
    Super Member Quacky's Avatar
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    Yuck. You can certainly take out a factor of a.
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  3. #3
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    Rationalize

    Rationalize before differentiating:

    \begin{array}{l}<br />
 \theta  = \frac{{1 - a\sqrt[3]{{{t^2}}}}}{{1 + a\sqrt {{t^3}} }} \cdot \frac{{1 - a\sqrt {{t^3}} }}{{1 - a\sqrt {{t^3}} }} \\ <br />
  = \frac{{\left( {1 - a\sqrt[3]{{{t^2}}}} \right)\left( {1 - a\sqrt {{t^3}} } \right)}}{{1 - {a^2}{t^3}}} \\ <br />
 \end{array}
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  4. #4
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    Quote Originally Posted by Quacky View Post


    Yuck. You can certainly take out a factor of a.
    I did think about factoring out an "a" but both terms in the denominator don't contain an "a", so I couldn't cancel out anything, right?
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  5. #5
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    Quote Originally Posted by hameerabbasi View Post
    Rationalize before differentiating:

    \begin{array}{l}<br />
 \theta  = \frac{{1 - a\sqrt[3]{{{t^2}}}}}{{1 + a\sqrt {{t^3}} }} \cdot \frac{{1 - a\sqrt {{t^3}} }}{{1 - a\sqrt {{t^3}} }} \\ <br />
  = \frac{{\left( {1 - a\sqrt[3]{{{t^2}}}} \right)\left( {1 - a\sqrt {{t^3}} } \right)}}{{1 - {a^2}{t^3}}} \\ <br />
 \end{array}
    What benefit do you see from rationalizing it first, it still ends up a pretty sloppy mess, unless I'm missing something?
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  6. #6
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    What I'm really after is how do you get from here:

    <br /> <br />
\frac{(1+at^{\frac{3}{2}})(\frac{-2}{3}at^{\frac{-1}{3}}) - (1-at^{\frac{2}{3}})(\frac{3}{2}at^{\frac{1}{2}})}{(1  +at^{\frac{3}{2}})^2}<br />

    to here:


    \frac{5a^2\sqrt[6]{t^7} - \frac{4a}{\sqrt[3]{t}} - 9a\sqrt{t}}{6(1+a\sqrt{t^3})^2}

    any help will be very much appreciated. Thanks!
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  7. #7
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    Multiply out the numerator, combine like terms, take out the common factor of 1/6 by putting 6 in the denominator.
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  8. #8
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    Quote Originally Posted by icemanfan View Post
    Multiply out the numerator, combine like terms, take out the common factor of 1/6 by putting 6 in the denominator.
    I did multiply out the numerator...look above. That is where I am stuck. I know the process, I was just wondering if someone could help "baby-step" through this one?
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  9. #9
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    Quote Originally Posted by dbakeg00 View Post
    I did multiply out the numerator...look above. That is where I am stuck. I know the process, I was just wondering if someone could help "baby-step" through this one?
    You made an error when multiplying out the numerator.

    \frac{-2}{3}at^\frac{-1}{3} + \frac{-2}{3}a^2t^{\frac{7}{6}} - \frac{3}{2}at^{\frac{1}{2}} - \frac{3}{2}a^2t^{\frac{7}{6}}

    should be
    \frac{-2}{3}at^\frac{-1}{3} + \frac{-2}{3}a^2t^{\frac{7}{6}} - \frac{3}{2}at^{\frac{1}{2}} + \frac{3}{2}a^2t^{\frac{7}{6}}.

    Then you can combine to get
    \left(\frac{3}{2} - \frac{2}{3}\right)a^2t^{\frac{7}{6}} - \frac{2}{3}at^\frac{-1}{3} - \frac{3}{2}at^{\frac{1}{2}}.

    Simplifying and using a common denominator of 6:
    \frac{5}{6}a^2t^{\frac{7}{6}} - \frac{4}{6}at^\frac{-1}{3} - \frac{9}{6}at^{\frac{1}{2}}.

    Putting the factor of 6 in the denominator leaves you with
    5a^2t^{\frac{7}{6}} - 4at^\frac{-1}{3} - 9at^{\frac{1}{2}} in the numerator.
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  10. #10
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    Thank you so much. That algebra error I made above was driving me nuts bc everything was not working out like I knew it should. Thank again for taking the time to help me.
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