when you post something there is an option which says'FORMATTING GUIDE' click on that and than
below it you'll see an option of 'EASY MATH EDITOR' click on that and than it will take to an app through which you can write like this.(use google chrome)

Hey buddy listen up! This sum is quite an easy one but your solution has made it far too complex. Simply take y=x^x^x and then take log on both sides then simply differentiate and voila! You will get your result in a jiffy by applying the product rule.

$</code> ... <code>$</code>...<code>."> Easy Math Editor

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## Comments

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TopNewest$let\quad \\ \quad { x }^{ { x }^{ x } }=y\quad \quad AND\quad { x }^{ x }=z\\ now\quad let\quad us\quad find\quad the\quad value\quad of\quad \frac { dz }{ dx\\ } \\ xlogx=logz\\ \frac { d }{ dx } xlogx=\frac { d }{ dx } logz\\ applying\quad the\quad product\quad rule\\ \frac { x }{ x } \quad +logx(1)=\frac { \frac { dz }{ dx } }{ z } \\ { x }^{ x }(1+logx)=\frac { dz }{ dx } \quad .........1.\\ \\ now\quad let\quad us\quad find\quad the\quad value\quad of\quad \frac { dy }{ dx } \\ { x }^{ x }logx=logy\\ \frac { d }{ dx } { x }^{ x }logx=\frac { d }{ dx } logy\\ applying\quad he\quad product\quad rule\\ \frac { { x }^{ x } }{ x } \quad +\quad logx(\frac { d }{ dx } { x }^{ x })=\frac { 1 }{ y } (\frac { dy }{ dx } )\\ \frac { { x }^{ x } }{ x } +\quad logx({ x }^{ x })(1\quad +\quad logx)=\frac { 1 }{ y } (\frac { dy }{ dx } )\\ \\ y(\frac { { x }^{ x } }{ x } \quad +\quad logx({ x }^{ x })(1\quad +\quad logx))=\frac { dy }{ dx } \\ ({ x }^{ { x }^{ x } })({ x }^{ x })(\frac { 1 }{ x } \quad +\quad logx(1\quad +\quad logx)=\frac { dy }{ dx } \\$

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How will I post my answer in this format? Please help me. I can't post my solution.

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when you post something there is an option which says'FORMATTING GUIDE' click on that and than below it you'll see an option of 'EASY MATH EDITOR' click on that and than it will take to an app through which you can write like this.(use google chrome)

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@Pranjal Jain @Calvin Lin @Chew-Seong Cheong @Yan Yau Cheng @math man @Krishna Sharma @Krishna Ar @abdulrahman khaled @Mehul Chaturvedi Upload solution

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But I have a solution. So I'm posting a new note in my profile.

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Take log and differentiate and for x^x again take log and differentiate

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Hey buddy listen up! This sum is quite an easy one but your solution has made it far too complex. Simply take y=x^x^x and then take log on both sides then simply differentiate and voila! You will get your result in a jiffy by applying the product rule.

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