# A Differentiation problem

Differentiate

$\huge x^{x^{x}}$ Note by Mehul Chaturvedi
5 years, 10 months ago

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

- 5 years, 10 months ago

- 5 years, 10 months ago

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)

- 5 years, 10 months ago

But u should use latex only for maths symbols

- 5 years, 10 months ago

yes i've got this....but i cant post the picture or the text...

- 5 years, 10 months ago

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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.

- 5 years, 9 months ago