anonymous
  • anonymous
help me pls find Dx [x^x^x^x^x]
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
find \[Dx [x ^{x ^{x ^{x ^{x}}}}]\]
wolfe8
  • wolfe8
Oh my. Is that really how the question looks like? If so, sorry but my brain rejects it at the sight.
anonymous
  • anonymous
@wolfe8 it's ok :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
lol i agree wolfe
wolfe8
  • wolfe8
Idk does it like involve log or ln or something
anonymous
  • anonymous
@fgiancarelli help
John_ES
  • John_ES
For the first iteration, you have, \[y=x^{x^{x^{x^x}}} \Rightarrow\ln y=x^{x^{x^{x}}}\ln x\] So, doing derivatives in both members, \[y'/y=(x^{x^{x^{x}}})'\ln x+x^{x^{x^{x}}}/x\] So, you see that you now will need the derivative of another function. To simplify, lets put the following nomenclature, \[y_1=x^{x^{x^{x^x}}}\]\[y_2=x^{x^{x^{x}}}\]\[y_3=x^{x^{x}}\]\[y_4=x^{x}\] And we proceed with the last one, \[y'/y_4=\ln x+1\Rightarrow y'=y_4(\ln x+1)\] Following the reasoning (from y_4 to y_1) you'll have, \[y'=y_1(y_2(y_3y_4(\ln x+1)\ln x+x^x/x)\ln x+x^{x^{x}}/x)\ln x+x^{x^{x^{x}}}/x\]
anonymous
  • anonymous
Personally I wouldn't do the nomenclature :P .

Looking for something else?

Not the answer you are looking for? Search for more explanations.