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



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Jonask
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\[(fg)'=f'g+fg'\]

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\[(2\sqrt{x})'\tan^{1}x+2\sqrt{x}(\tan^{1}x)'\]

Jonask
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the sqrt{x} is left out for the arctan ,mistake

Study23
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Okay. So what do I do next?????

Jonask
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' means derive

Study23
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So what do I do after this?dw:1352961580767:dw

Jonask
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done

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dw:1353055292887:dw

Study23
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Thanks for pointing that out! And thank you for your help!

Jonask
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dw:1353055340810:dw

Study23
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HOw?

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How is it ()?

Jonask
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\[(\sqrt{x})'=\frac{1}{2\sqrt{x}}\]

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Oh I see now!

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Power Rule!

Jonask
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notice that for \[(\tan^{1}\sqrt{x})'=(\sqrt{x})'(\frac{ 1 }{ 1+(\sqrt{x})^2 })\]
\[\frac{ (1/2)x^{1/2} }{ (1+x) }\]