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

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

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

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

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

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

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

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

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

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

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

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

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

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

Jonask Group TitleBest ResponseYou've already chosen the best response.3
notice that for \[(\tan^{1}\sqrt{x})'=(\sqrt{x})'(\frac{ 1 }{ 1+(\sqrt{x})^2 })\] \[\frac{ (1/2)x^{1/2} }{ (1+x) }\]
 2 years ago
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