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

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1. anonymous

$y=(\ln x)^{\sqrt{x}}$

2. zepdrix

Hey Jess :) We can uhhh.. use logs I suppose, that should help

3. zepdrix

So umm.. ya ya let's try that. Let's take the natural log of each side.$\large\rm \ln y= \ln\left[(\ln x)^{\sqrt x}\right]$

4. zepdrix

Now 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

5. anonymous

im somewhat getting it

6. zepdrix

So 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.

7. zepdrix

Product rule on the right, ya?

8. zepdrix

Too much? :O Brain esplode?

9. anonymous

no im still here

10. zepdrix

So then uhhh take some derivatives :) remember derivative of $$\large\rm \sqrt{x}$$ ?

11. nincompoop