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Study23
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Find the derivative?
\(\ \Large 2\sqrt{x}tan^{1}\sqrt{x} \)
 one year ago
 one year ago
Study23 Group Title
Find the derivative? \(\ \Large 2\sqrt{x}tan^{1}\sqrt{x} \)
 one year ago
 one year ago

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Jonask Group TitleBest ResponseYou've already chosen the best response.3
\[(fg)'=f'g+fg'\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.3
\[(2\sqrt{x})'\tan^{1}x+2\sqrt{x}(\tan^{1}x)'\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.3
the sqrt{x} is left out for the arctan ,mistake
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Okay. So what do I do next?????
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.3
' means derive
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
So what do I do after this?dw:1352961580767:dw
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.3
dw:1353055292887:dw
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Thanks for pointing that out! And thank you for your help!
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.3
dw:1353055340810:dw
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
How is it ()?
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.3
\[(\sqrt{x})'=\frac{1}{2\sqrt{x}}\]
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Oh I see now!
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Power Rule!
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.3
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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