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Math Help - Derivation of this function..?

  1. #1
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    Derivation of this function..?

    Hello.Nice to meet you math helper's.I'm new here sorry if i accideantly broke any rule of this forum.How i underdstand for any quesiton of derivation i can ask in this section of this forum.Sorry for my bad english :$..Hope that we will spread our knowledge.
    I insert as a picture my two problems..
    Maybe for someone is very easy for me are extremly hard..

    Thnx and sorry..
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  2. #2
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    Hello, TrumBeta!


    Welcome aboard!

    I must assume you know the Quotient Rule: . \left(\frac{u}{v}\right)' \;=\;\frac{v\!\cdot\! u' - u\!\cdot\! v'}{v^2}


    y \;=\;\frac{1+x\sqrt{x}}{1 - x\sqrt{x}}
    We have: . y \;=\;\frac{1 + x^{\frac{3}{2}}}{1 - x^{\frac{3}{2}}}

    Then: . y' \;=\;\frac{\left(1-x^{\frac{3}{2}}\right) \left(\frac{3}{2}x^{\frac{1}{2}}\right) - \left(1+x^{\frac{3}{2}}\right)\left(-\frac{3}{2}x^{\frac{1}{2}}\right)} {\left(1 - x^{\frac{3}{2}}\right)^2}

    . . . . . . . = \;\frac{\frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^2 + \frac{3}{2}x^{\frac{1}{2}} + \frac{3}{2}x^2}{\left(1-x^{\frac{3}{2}}\right)^2}

    . . . . . . . = \;\frac{3x^{\frac{1}{2}}}{\left(1-x^{\frac{3}{2}}\right)^2}



    y \;=\;\frac{e^x + e^{-x}}{e^x - e^{-x}}

    y' \;=\;\frac{(e^x-e^{-x})(e^x-e^{-x}) - (e^x + e^{-x})(e^x+e^{-x})} {(e^x-e^{-x})^2}

    . . =\; \frac{\left(e^{2x} - 1 - 1 + e^{-2x}\right) - \left(e^{2x} + 1 + 1 + e^{-2x}\right)}{\left(e^x-e^{-x}\right)^2}<br />

    . . =\; \frac{e^{2x} - 2 + e^{-2x} - e^{2x} - 2 - e^{-2x}}{(e^x - e^{-x})^2}

    . . =\; \frac{-4}{(e^x-e^{-x})^2}

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  3. #3
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    thank you very much .. you helped me a lot...!
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