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Study23 Group Title

Find the derivative? \(\ \Large 2\sqrt{x}tan^{-1}\sqrt{x} \)

  • one year ago
  • one year ago

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  1. Jonask Group Title
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    \[(fg)'=f'g+fg'\]

    • one year ago
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    \[(2\sqrt{x})'\tan^{-1}x+2\sqrt{x}(\tan^{-1}x)'\]

    • one year ago
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    the sqrt{x} is left out for the arctan ,mistake

    • one year ago
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    Okay. So what do I do next?????

    • one year ago
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    ' means derive

    • one year ago
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    So what do I do after this?|dw:1352961580767:dw|

    • one year ago
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    done

    • one year ago
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    |dw:1353055292887:dw|

    • one year ago
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    Thanks for pointing that out! And thank you for your help!

    • one year ago
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    |dw:1353055340810:dw|

    • one year ago
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    HOw?

    • one year ago
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    How is it (-)?

    • one year ago
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    \[(\sqrt{x})'=\frac{1}{2\sqrt{x}}\]

    • one year ago
  14. Study23 Group Title
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    Oh I see now!

    • one year ago
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    Power Rule!

    • one year ago
  16. Jonask Group Title
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    notice that for \[(\tan^{-1}\sqrt{x})'=(\sqrt{x})'(\frac{ 1 }{ 1+(\sqrt{x})^2 })\] \[\frac{ (1/2)x^{-1/2} }{ (1+x) }\]

    • one year ago
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