anonymous
  • anonymous
Find the derivative? \(\ \Large 2\sqrt{x}tan^{-1}\sqrt{x} \)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[(fg)'=f'g+fg'\]
anonymous
  • anonymous
\[(2\sqrt{x})'\tan^{-1}x+2\sqrt{x}(\tan^{-1}x)'\]
anonymous
  • anonymous
the sqrt{x} is left out for the arctan ,mistake

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anonymous
  • anonymous
Okay. So what do I do next?????
anonymous
  • anonymous
' means derive
anonymous
  • anonymous
So what do I do after this?|dw:1352961580767:dw|
anonymous
  • anonymous
done
anonymous
  • anonymous
|dw:1353055292887:dw|
anonymous
  • anonymous
Thanks for pointing that out! And thank you for your help!
anonymous
  • anonymous
|dw:1353055340810:dw|
anonymous
  • anonymous
HOw?
anonymous
  • anonymous
How is it (-)?
anonymous
  • anonymous
\[(\sqrt{x})'=\frac{1}{2\sqrt{x}}\]
anonymous
  • anonymous
Oh I see now!
anonymous
  • anonymous
Power Rule!
anonymous
  • anonymous
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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