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

Mathematics
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\[(fg)'=f'g+fg'\]
\[(2\sqrt{x})'\tan^{-1}x+2\sqrt{x}(\tan^{-1}x)'\]
the sqrt{x} is left out for the arctan ,mistake

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Other answers:

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