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Math Help - Differential

  1. #1
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    Differential

    Hello,

    I have here a strength differential \frac{-t}{\sqrt{1-t^2}}

    Normally is not a big problem but I making a mistake I don't know:

    let my try u=1-t^2 thus u'=-2t

    v=\sqrt{u} thus v'=\frac{1}{2}u^{-\frac{1}{2}

    And w=\frac{-t}{v} thus w'=\frac{-v+tv'}{v^2}

    Now I need to multiply al the thinks and then I find \frac{t}{1-t^2} but it's not correct.

    Where do I miss ? Greets.
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  2. #2
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    Quote Originally Posted by Bert
    Hello,

    I have here a strength differential \frac{-t}{\sqrt{1-t^2}}

    Normally is not a big problem but I making a mistake I don't know:

    let my try u=1-t^2 thus u'=-2t

    v=\sqrt{u} thus v'=\frac{1}{2}u^{-\frac{1}{2}

    And w=\frac{-t}{v} thus w'=\frac{-v+tv'}{v^2}

    Now I need to multiply al the thinks and then I find \frac{t}{1-t^2} but it's not correct.

    Where do I miss ? Greets.
    Stage 1 - Product rule:

    <br />
\frac{d}{dt}\left(\frac{-t}{\sqrt{1-t^2}}\right)=\frac{1}{\sqrt{1-t^2}}\frac{d}{dt} (-t)+(-t)\frac{d}{dt} \ \left(\frac{1}{\sqrt{1-t^2}}\right)<br />
    <br />
=-\frac{1}{\sqrt{1-t^2}}-t\ \frac{d}{dt} \ \left(\frac{1}{\sqrt{1-t^2}}\right)<br />

    Stage 2 - Chain rule on last term on the right:


    <br />
=-\frac{1}{\sqrt{1-t^2}}-t\ (-1/2)\frac{1}{(1-t^2)^{3/2}}\frac{d}{dt}(1-t^2)<br />

    <br />
=\frac{1}{\sqrt{1-t^2}}-t\ (-1/2)\frac{1}{(1-t^2)^{3/2}}(-2t)<br />

    <br />
=\frac{1}{\sqrt{1-t^2}}-\frac{t^2}{(1-t^2)^{3/2}}<br />

    From here you should be able to simplify this yourself.

    RonL
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  3. #3
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    thus you can splits this in two parts by the produc rule

    then you have \frac{1}{\sqrt{1-t^2}}

    I can find then on this way: u=1-t^2 Thus u'=-2t

    Then v=\sqrt{u} and v'=\frac{1}{2}u^{-\frac{1}{2}}

    Then \frac{\frac{-t}{\sqrt{1-t^2}}}{\sqrt{1-t^2}^2}=\frac{-t}{\sqrt{1-t^2}}\frac{1}{1-t^2}

    But why is this fault ? u=1-t^2 \rightarrow u'=-2t \ \ \ v=\sqrt{u} \rightarrow\frac{1}{2}u^{-\frac{1}{2}} \ \ \ w=\frac{1}{v} \rightarrow w'=lnv

    And then multiply them what is mis? Greets.
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  4. #4
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    Quote Originally Posted by Bert
    But why is this fault ? u=1-t^2 \rightarrow u'=-2t \ \ \ v=\sqrt{u} \rightarrow\frac{1}{2}u^{-\frac{1}{2}} \ \ \ w=\frac{1}{v} \rightarrow w'=lnv

    And then multiply them what is mis? Greets.
    Mistake that if,
    w=\frac{1}{v}
    then,
    w'=-\frac{1}{v^2}
    You confused with anti-derivative.
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  5. #5
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    of cours thank you very well.

    a small problem If I simplify \frac{-t}{\sqrt{1-t^2}}\frac{1}{1-t^2}<br />

    then I get -\frac{t}{{(1-t^2)}^{\frac{3}{2}}}} And this it not the same as I differentiate the original by my computer program then I get -\frac{1}{{(1-t^2)}^\frac{3}{2}}
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  6. #6
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    Quote Originally Posted by Bert
    thus you can splits this in two parts by the produc rule
    then you have \frac{1}{\sqrt{1-t^2}}
    I can find then on this way: u=1-t^2 Thus u'=-2t

    Then v=\sqrt{u} and v'=\frac{1}{2}u^{-\frac{1}{2}}

    ...

    Hello,

    Then v=\sqrt{u} and v'=\frac{1}{2}u^{-\frac{1}{2}}\ \gets \mbox{this looks funny to me}

    If you use the chain rule on v' then you get: v'=\frac{1}{2}u^{-\frac{1}{2}} \cdot (-2t)=-t \cdot u^{-\frac{1}{2}}

    Greetings

    EB
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  7. #7
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    Thank you very well I find them great good forum Greets Thanks
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