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Study23

  • 2 years ago

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

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  1. Jonask
    • 2 years ago
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    \[(fg)'=f'g+fg'\]

  2. Jonask
    • 2 years ago
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    \[(2\sqrt{x})'\tan^{-1}x+2\sqrt{x}(\tan^{-1}x)'\]

  3. Jonask
    • 2 years ago
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    the sqrt{x} is left out for the arctan ,mistake

  4. Study23
    • 2 years ago
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    Okay. So what do I do next?????

  5. Jonask
    • 2 years ago
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    ' means derive

  6. Study23
    • 2 years ago
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    So what do I do after this?|dw:1352961580767:dw|

  7. Jonask
    • 2 years ago
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    done

  8. Jonask
    • 2 years ago
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    |dw:1353055292887:dw|

  9. Study23
    • 2 years ago
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    Thanks for pointing that out! And thank you for your help!

  10. Jonask
    • 2 years ago
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    |dw:1353055340810:dw|

  11. Study23
    • 2 years ago
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    HOw?

  12. Study23
    • 2 years ago
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    How is it (-)?

  13. Jonask
    • 2 years ago
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    \[(\sqrt{x})'=\frac{1}{2\sqrt{x}}\]

  14. Study23
    • 2 years ago
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    Oh I see now!

  15. Study23
    • 2 years ago
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    Power Rule!

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

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