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TopNewest\(let\quad \\ \quad { x }^{ { x }^{ x } }=y\quad \quad AND\quad { x }^{ x }=z\\ now\quad let\quad us\quad find\quad the\quad value\quad of\quad \frac { dz }{ dx\\ } \\ xlogx=logz\\ \frac { d }{ dx } xlogx=\frac { d }{ dx } logz\\ applying\quad the\quad product\quad rule\\ \frac { x }{ x } \quad +logx(1)=\frac { \frac { dz }{ dx } }{ z } \\ { x }^{ x }(1+logx)=\frac { dz }{ dx } \quad .........1.\\ \\ now\quad let\quad us\quad find\quad the\quad value\quad of\quad \frac { dy }{ dx } \\ { x }^{ x }logx=logy\\ \frac { d }{ dx } { x }^{ x }logx=\frac { d }{ dx } logy\\ applying\quad he\quad product\quad rule\\ \frac { { x }^{ x } }{ x } \quad +\quad logx(\frac { d }{ dx } { x }^{ x })=\frac { 1 }{ y } (\frac { dy }{ dx } )\\ \frac { { x }^{ x } }{ x } +\quad logx({ x }^{ x })(1\quad +\quad logx)=\frac { 1 }{ y } (\frac { dy }{ dx } )\\ \\ y(\frac { { x }^{ x } }{ x } \quad +\quad logx({ x }^{ x })(1\quad +\quad logx))=\frac { dy }{ dx } \\ ({ x }^{ { x }^{ x } })({ x }^{ x })(\frac { 1 }{ x } \quad +\quad logx(1\quad +\quad logx)=\frac { dy }{ dx } \\ \) – Prashant Ramnani · 2 years, 2 months ago

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– Trishit Chandra · 2 years, 2 months ago

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

– Prashant Ramnani · 2 years, 2 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)Log in to reply

– Trishit Chandra · 2 years, 2 months ago

yes i've got this....but i cant post the picture or the text...Log in to reply

– Mehul Chaturvedi · 2 years, 2 months ago

But u should use latex only for maths symbolsLog in to reply

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– Trishit Chandra · 2 years, 2 months ago

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

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, 1 month ago

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Take log and differentiate and for x^x again take log and differentiate – Rohit Shah · 2 years, 2 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, 2 months ago

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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 – Mehul Chaturvedi · 2 years, 2 months ago

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