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A Differentiation problem

Differentiate

\[\huge x^{x^{x}}\]

Note by Mehul Chaturvedi
2 years, 9 months ago

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

Prashant Ramnani - 2 years, 9 months ago

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

Trishit Chandra - 2 years, 9 months ago

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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)

Prashant Ramnani - 2 years, 9 months ago

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

Trishit Chandra - 2 years, 9 months ago

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@Prashant Ramnani But u should use latex only for maths symbols

Mehul Chaturvedi - 2 years, 9 months ago

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Comment deleted Jan 13, 2015

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no.... the answer is (x^x^x)[log(x^x^x)(logx+1)+x^(x-1)]

Trishit Chandra - 2 years, 9 months ago

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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.

Srutarshi Chakrabarti - 2 years, 8 months ago

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

Rohit Shah - 2 years, 9 months ago

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

Trishit Chandra - 2 years, 9 months ago

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