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anonymous
 one year ago
differentiate y=(Inx)^squart root x Do not simplify answer
anonymous
 one year ago
differentiate y=(Inx)^squart root x Do not simplify answer

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=(\ln x)^{\sqrt{x}}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Hey Jess :) We can uhhh.. use logs I suppose, that should help

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So umm.. ya ya let's try that. Let's take the natural log of each side.\[\large\rm \ln y= \ln\left[(\ln x)^{\sqrt x}\right]\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Now we can apply our log rule to rewrite the right side,\[\large\rm \ln y=\sqrt{x} ~\ln\left[(\ln x)\right]\]And from there you'll apply product rule :) It's gonna get messy! Confused by any of that? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im somewhat getting it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So let's first `set up` our derivative so we understand how it's going to look.\[\large\rm \color{royalblue}{(\ln y)'}=\color{royalblue}{(\sqrt x)'}\ln(\ln x)+\sqrt x\color{royalblue}{\left[\ln(\ln x)\right]'}\]The blue stuff is what we need to differentiate.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Product rule on the right, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Too much? :O Brain esplode?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So then uhhh take some derivatives :) remember derivative of \(\large\rm \sqrt{x}\) ?

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/559c9c4ae4b0564dd2d41b6a
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