## Study23 Group Title Find the derivative? $$\ \Large 2\sqrt{x}tan^{-1}\sqrt{x}$$ 2 years ago 2 years ago

$(fg)'=f'g+fg'$

$(2\sqrt{x})'\tan^{-1}x+2\sqrt{x}(\tan^{-1}x)'$

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

4. Study23

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

' means derive

6. Study23

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

done

|dw:1353055292887:dw|

9. Study23

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

|dw:1353055340810:dw|

11. Study23

HOw?

12. Study23

How is it (-)?

$(\sqrt{x})'=\frac{1}{2\sqrt{x}}$

14. Study23

Oh I see now!

15. Study23

Power Rule!

notice that for $(\tan^{-1}\sqrt{x})'=(\sqrt{x})'(\frac{ 1 }{ 1+(\sqrt{x})^2 })$ $\frac{ (1/2)x^{-1/2} }{ (1+x) }$