Problem with derivative of $x^{x^x}$

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I was recently watching blackpenredpen’s video (found here: https://m.youtube.com/watch?v=UJ3Ahpcvmf8) where he found the derivative of the the function $y = x^{x^x}$. Before watching the video, I decided to try it myself, and I didn’t it this way: Firstly I decided I would use the chain rule and define two functions, $u(x)$ and $v(x)$, both of which are $x^x$, so I could show the function as $u(v(x))$. Using the chain rule, the derivative would be: $u’(v(x))cdot v’(x)$. We can find the derivative of $x^x$ (as shown on Quora here: https://www.quora.com/What-is-the-derivative-of-x-x-How-could-I-derive-it-as-well) as $x^xcdot(ln x+1)$. Now using the chain rule to get $u’(v(x))$ we get ${(x^x)}^{(x^x)}cdot(ln {x^x}+1)$ or $x^{x^{x+1}}cdot(ln {x^x}+1)$. This wasn’t the answer that blackpenredpen arrived at though. In the video, he arrived at $x^xcdot x^{x^x}left({frac{1}{x}+ln x+(ln x)^2}right)$. I can’t see how he did it, and I’m sure that it is right. If anyone could confirm my answer or show where I went wrong it would be greatly appreciated.
Thanks.










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  • 2




    note that $u(v(x))=v(x)^{v(x)}=(x^x)^{x^x}$, what is not the same function. Instead try the following: $x^{(x^x)}=exp(x^xln x)=exp(e^{xln x}ln x)$, now use the chain rule.
    – Masacroso
    Nov 15 at 6:10












  • @Masacroso Are you sure? Exponentiation is left associative so $alpha^{alpha^alpha}$ = $alpha^{(alpha^alpha)}$. So wouldn’t ${(x^x)}^{x^x} = {(x^x)}^{(x^x)}$?
    – L. McDonald
    Nov 15 at 6:19












  • Yes, that is right. But the way I (tried) to find $frac{d}{dx}left[{x^{x^x}}right]$ lead me to get a ${(x^x)}^{(x^x)}$ in my answer, this is probably where I’ve gone wrong. Note: this was in response to another comment, which seems to have been deleted.
    – L. McDonald
    Nov 15 at 6:28

















up vote
0
down vote

favorite












I was recently watching blackpenredpen’s video (found here: https://m.youtube.com/watch?v=UJ3Ahpcvmf8) where he found the derivative of the the function $y = x^{x^x}$. Before watching the video, I decided to try it myself, and I didn’t it this way: Firstly I decided I would use the chain rule and define two functions, $u(x)$ and $v(x)$, both of which are $x^x$, so I could show the function as $u(v(x))$. Using the chain rule, the derivative would be: $u’(v(x))cdot v’(x)$. We can find the derivative of $x^x$ (as shown on Quora here: https://www.quora.com/What-is-the-derivative-of-x-x-How-could-I-derive-it-as-well) as $x^xcdot(ln x+1)$. Now using the chain rule to get $u’(v(x))$ we get ${(x^x)}^{(x^x)}cdot(ln {x^x}+1)$ or $x^{x^{x+1}}cdot(ln {x^x}+1)$. This wasn’t the answer that blackpenredpen arrived at though. In the video, he arrived at $x^xcdot x^{x^x}left({frac{1}{x}+ln x+(ln x)^2}right)$. I can’t see how he did it, and I’m sure that it is right. If anyone could confirm my answer or show where I went wrong it would be greatly appreciated.
Thanks.










share|cite|improve this question




















  • 2




    note that $u(v(x))=v(x)^{v(x)}=(x^x)^{x^x}$, what is not the same function. Instead try the following: $x^{(x^x)}=exp(x^xln x)=exp(e^{xln x}ln x)$, now use the chain rule.
    – Masacroso
    Nov 15 at 6:10












  • @Masacroso Are you sure? Exponentiation is left associative so $alpha^{alpha^alpha}$ = $alpha^{(alpha^alpha)}$. So wouldn’t ${(x^x)}^{x^x} = {(x^x)}^{(x^x)}$?
    – L. McDonald
    Nov 15 at 6:19












  • Yes, that is right. But the way I (tried) to find $frac{d}{dx}left[{x^{x^x}}right]$ lead me to get a ${(x^x)}^{(x^x)}$ in my answer, this is probably where I’ve gone wrong. Note: this was in response to another comment, which seems to have been deleted.
    – L. McDonald
    Nov 15 at 6:28















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was recently watching blackpenredpen’s video (found here: https://m.youtube.com/watch?v=UJ3Ahpcvmf8) where he found the derivative of the the function $y = x^{x^x}$. Before watching the video, I decided to try it myself, and I didn’t it this way: Firstly I decided I would use the chain rule and define two functions, $u(x)$ and $v(x)$, both of which are $x^x$, so I could show the function as $u(v(x))$. Using the chain rule, the derivative would be: $u’(v(x))cdot v’(x)$. We can find the derivative of $x^x$ (as shown on Quora here: https://www.quora.com/What-is-the-derivative-of-x-x-How-could-I-derive-it-as-well) as $x^xcdot(ln x+1)$. Now using the chain rule to get $u’(v(x))$ we get ${(x^x)}^{(x^x)}cdot(ln {x^x}+1)$ or $x^{x^{x+1}}cdot(ln {x^x}+1)$. This wasn’t the answer that blackpenredpen arrived at though. In the video, he arrived at $x^xcdot x^{x^x}left({frac{1}{x}+ln x+(ln x)^2}right)$. I can’t see how he did it, and I’m sure that it is right. If anyone could confirm my answer or show where I went wrong it would be greatly appreciated.
Thanks.










share|cite|improve this question















I was recently watching blackpenredpen’s video (found here: https://m.youtube.com/watch?v=UJ3Ahpcvmf8) where he found the derivative of the the function $y = x^{x^x}$. Before watching the video, I decided to try it myself, and I didn’t it this way: Firstly I decided I would use the chain rule and define two functions, $u(x)$ and $v(x)$, both of which are $x^x$, so I could show the function as $u(v(x))$. Using the chain rule, the derivative would be: $u’(v(x))cdot v’(x)$. We can find the derivative of $x^x$ (as shown on Quora here: https://www.quora.com/What-is-the-derivative-of-x-x-How-could-I-derive-it-as-well) as $x^xcdot(ln x+1)$. Now using the chain rule to get $u’(v(x))$ we get ${(x^x)}^{(x^x)}cdot(ln {x^x}+1)$ or $x^{x^{x+1}}cdot(ln {x^x}+1)$. This wasn’t the answer that blackpenredpen arrived at though. In the video, he arrived at $x^xcdot x^{x^x}left({frac{1}{x}+ln x+(ln x)^2}right)$. I can’t see how he did it, and I’m sure that it is right. If anyone could confirm my answer or show where I went wrong it would be greatly appreciated.
Thanks.







calculus derivatives chain-rule tetration






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edited Nov 15 at 7:51

























asked Nov 15 at 6:03









L. McDonald

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  • 2




    note that $u(v(x))=v(x)^{v(x)}=(x^x)^{x^x}$, what is not the same function. Instead try the following: $x^{(x^x)}=exp(x^xln x)=exp(e^{xln x}ln x)$, now use the chain rule.
    – Masacroso
    Nov 15 at 6:10












  • @Masacroso Are you sure? Exponentiation is left associative so $alpha^{alpha^alpha}$ = $alpha^{(alpha^alpha)}$. So wouldn’t ${(x^x)}^{x^x} = {(x^x)}^{(x^x)}$?
    – L. McDonald
    Nov 15 at 6:19












  • Yes, that is right. But the way I (tried) to find $frac{d}{dx}left[{x^{x^x}}right]$ lead me to get a ${(x^x)}^{(x^x)}$ in my answer, this is probably where I’ve gone wrong. Note: this was in response to another comment, which seems to have been deleted.
    – L. McDonald
    Nov 15 at 6:28
















  • 2




    note that $u(v(x))=v(x)^{v(x)}=(x^x)^{x^x}$, what is not the same function. Instead try the following: $x^{(x^x)}=exp(x^xln x)=exp(e^{xln x}ln x)$, now use the chain rule.
    – Masacroso
    Nov 15 at 6:10












  • @Masacroso Are you sure? Exponentiation is left associative so $alpha^{alpha^alpha}$ = $alpha^{(alpha^alpha)}$. So wouldn’t ${(x^x)}^{x^x} = {(x^x)}^{(x^x)}$?
    – L. McDonald
    Nov 15 at 6:19












  • Yes, that is right. But the way I (tried) to find $frac{d}{dx}left[{x^{x^x}}right]$ lead me to get a ${(x^x)}^{(x^x)}$ in my answer, this is probably where I’ve gone wrong. Note: this was in response to another comment, which seems to have been deleted.
    – L. McDonald
    Nov 15 at 6:28










2




2




note that $u(v(x))=v(x)^{v(x)}=(x^x)^{x^x}$, what is not the same function. Instead try the following: $x^{(x^x)}=exp(x^xln x)=exp(e^{xln x}ln x)$, now use the chain rule.
– Masacroso
Nov 15 at 6:10






note that $u(v(x))=v(x)^{v(x)}=(x^x)^{x^x}$, what is not the same function. Instead try the following: $x^{(x^x)}=exp(x^xln x)=exp(e^{xln x}ln x)$, now use the chain rule.
– Masacroso
Nov 15 at 6:10














@Masacroso Are you sure? Exponentiation is left associative so $alpha^{alpha^alpha}$ = $alpha^{(alpha^alpha)}$. So wouldn’t ${(x^x)}^{x^x} = {(x^x)}^{(x^x)}$?
– L. McDonald
Nov 15 at 6:19






@Masacroso Are you sure? Exponentiation is left associative so $alpha^{alpha^alpha}$ = $alpha^{(alpha^alpha)}$. So wouldn’t ${(x^x)}^{x^x} = {(x^x)}^{(x^x)}$?
– L. McDonald
Nov 15 at 6:19














Yes, that is right. But the way I (tried) to find $frac{d}{dx}left[{x^{x^x}}right]$ lead me to get a ${(x^x)}^{(x^x)}$ in my answer, this is probably where I’ve gone wrong. Note: this was in response to another comment, which seems to have been deleted.
– L. McDonald
Nov 15 at 6:28






Yes, that is right. But the way I (tried) to find $frac{d}{dx}left[{x^{x^x}}right]$ lead me to get a ${(x^x)}^{(x^x)}$ in my answer, this is probably where I’ve gone wrong. Note: this was in response to another comment, which seems to have been deleted.
– L. McDonald
Nov 15 at 6:28












2 Answers
2






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up vote
2
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You can apply the method given on quora to your problem:
$$left(x^{x^x}right)'=left(e^{x^xln x}right)'=e^{x^xln x}cdot left(x^xln xright)'=\
=e^{x^xln x}cdot left(e^{xln x}ln xright)'=\
=e^{x^xln x}cdot left[left(e^{xln x}right)'ln x+e^{xln x}cdot frac1xright]=\
=x^{x^x}cdot left[e^{xln x}left(xln xright)'ln x+x^xcdot frac1xright]=\
=x^{x^x}cdot left[x^xleft(ln x+1right)ln x+x^xcdot frac1xright]=\
=x^{x^x}cdot x^xleft[ln^2 x+ln x+frac1xright].$$

P.S. I could not watch the referenced video on Youtube, because it is not opening here (it might be blocked).






share|cite|improve this answer





















  • Sorry about the video, he did almost exactly the same thing that you did, but might I ask, would you be able to show me where I went wrong?
    – L. McDonald
    Nov 15 at 7:49










  • If $u(x)=x^x, v(x)=x^x$, then $u(v(x))=(x^x)^{x^x}=x^{x^{x+1}}$, which is a different function.
    – farruhota
    Nov 15 at 7:54










  • Oh, I see, $u(v(x))$ would have been $x^{x^x}$ of course. That’s a bit embarrassing...
    – L. McDonald
    Nov 15 at 7:56


















up vote
2
down vote













For differentiating expressions of the form $u(x)^{v(x)},$ I've always found it easiest to take logarithms and then differentiate implicitly. However, rather than work with the more general form $y = u^v,$ I'll work with $y = x^U$ (i.e. use $x$ for the base), which is sufficient to deal with $x^{x^x}.$



Taking the logarithm of both sides gives



$$ y = x^U ;; implies ln y = lnleft(x^U right) ;; implies ;; ln y = U cdot ln x $$



Now differentiate both sides with respect to $x$:



$$ frac{d}{dx}left(ln yright) ;; = ;; frac{d}{dx}left(U cdot ln x right) $$



$$ frac{y'}{y} ;; = ;; frac{d}{dx}(U) cdot ln x ; + ; U cdot frac{d}{dx} (ln x) ;; = ;; U'ln x ; + ; frac{U}{x} $$



$$ y' ;; = ;; y cdot left[ U'ln x ; + ; frac{U}{x} right] ;; = ;; x^U cdot left[ U'ln x ; + ; frac{U}{x} right] $$



Using this result for $U(x) = x,$ we get



$$ left(x^xright)' ;; = ;; x^x cdot left[ 1 cdot ln x ; + ; frac{x}{x} right] ;; = ;; x^x(ln x + 1) $$



Using that same result for $U(x) = x^x$ along with what we just found for $left(x^xright)',$ we get



$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$



$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$



$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$



$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$






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    2 Answers
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    2 Answers
    2






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    up vote
    2
    down vote



    accepted










    You can apply the method given on quora to your problem:
    $$left(x^{x^x}right)'=left(e^{x^xln x}right)'=e^{x^xln x}cdot left(x^xln xright)'=\
    =e^{x^xln x}cdot left(e^{xln x}ln xright)'=\
    =e^{x^xln x}cdot left[left(e^{xln x}right)'ln x+e^{xln x}cdot frac1xright]=\
    =x^{x^x}cdot left[e^{xln x}left(xln xright)'ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot left[x^xleft(ln x+1right)ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot x^xleft[ln^2 x+ln x+frac1xright].$$

    P.S. I could not watch the referenced video on Youtube, because it is not opening here (it might be blocked).






    share|cite|improve this answer





















    • Sorry about the video, he did almost exactly the same thing that you did, but might I ask, would you be able to show me where I went wrong?
      – L. McDonald
      Nov 15 at 7:49










    • If $u(x)=x^x, v(x)=x^x$, then $u(v(x))=(x^x)^{x^x}=x^{x^{x+1}}$, which is a different function.
      – farruhota
      Nov 15 at 7:54










    • Oh, I see, $u(v(x))$ would have been $x^{x^x}$ of course. That’s a bit embarrassing...
      – L. McDonald
      Nov 15 at 7:56















    up vote
    2
    down vote



    accepted










    You can apply the method given on quora to your problem:
    $$left(x^{x^x}right)'=left(e^{x^xln x}right)'=e^{x^xln x}cdot left(x^xln xright)'=\
    =e^{x^xln x}cdot left(e^{xln x}ln xright)'=\
    =e^{x^xln x}cdot left[left(e^{xln x}right)'ln x+e^{xln x}cdot frac1xright]=\
    =x^{x^x}cdot left[e^{xln x}left(xln xright)'ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot left[x^xleft(ln x+1right)ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot x^xleft[ln^2 x+ln x+frac1xright].$$

    P.S. I could not watch the referenced video on Youtube, because it is not opening here (it might be blocked).






    share|cite|improve this answer





















    • Sorry about the video, he did almost exactly the same thing that you did, but might I ask, would you be able to show me where I went wrong?
      – L. McDonald
      Nov 15 at 7:49










    • If $u(x)=x^x, v(x)=x^x$, then $u(v(x))=(x^x)^{x^x}=x^{x^{x+1}}$, which is a different function.
      – farruhota
      Nov 15 at 7:54










    • Oh, I see, $u(v(x))$ would have been $x^{x^x}$ of course. That’s a bit embarrassing...
      – L. McDonald
      Nov 15 at 7:56













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    You can apply the method given on quora to your problem:
    $$left(x^{x^x}right)'=left(e^{x^xln x}right)'=e^{x^xln x}cdot left(x^xln xright)'=\
    =e^{x^xln x}cdot left(e^{xln x}ln xright)'=\
    =e^{x^xln x}cdot left[left(e^{xln x}right)'ln x+e^{xln x}cdot frac1xright]=\
    =x^{x^x}cdot left[e^{xln x}left(xln xright)'ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot left[x^xleft(ln x+1right)ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot x^xleft[ln^2 x+ln x+frac1xright].$$

    P.S. I could not watch the referenced video on Youtube, because it is not opening here (it might be blocked).






    share|cite|improve this answer












    You can apply the method given on quora to your problem:
    $$left(x^{x^x}right)'=left(e^{x^xln x}right)'=e^{x^xln x}cdot left(x^xln xright)'=\
    =e^{x^xln x}cdot left(e^{xln x}ln xright)'=\
    =e^{x^xln x}cdot left[left(e^{xln x}right)'ln x+e^{xln x}cdot frac1xright]=\
    =x^{x^x}cdot left[e^{xln x}left(xln xright)'ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot left[x^xleft(ln x+1right)ln x+x^xcdot frac1xright]=\
    =x^{x^x}cdot x^xleft[ln^2 x+ln x+frac1xright].$$

    P.S. I could not watch the referenced video on Youtube, because it is not opening here (it might be blocked).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 15 at 7:11









    farruhota

    17.8k2736




    17.8k2736












    • Sorry about the video, he did almost exactly the same thing that you did, but might I ask, would you be able to show me where I went wrong?
      – L. McDonald
      Nov 15 at 7:49










    • If $u(x)=x^x, v(x)=x^x$, then $u(v(x))=(x^x)^{x^x}=x^{x^{x+1}}$, which is a different function.
      – farruhota
      Nov 15 at 7:54










    • Oh, I see, $u(v(x))$ would have been $x^{x^x}$ of course. That’s a bit embarrassing...
      – L. McDonald
      Nov 15 at 7:56


















    • Sorry about the video, he did almost exactly the same thing that you did, but might I ask, would you be able to show me where I went wrong?
      – L. McDonald
      Nov 15 at 7:49










    • If $u(x)=x^x, v(x)=x^x$, then $u(v(x))=(x^x)^{x^x}=x^{x^{x+1}}$, which is a different function.
      – farruhota
      Nov 15 at 7:54










    • Oh, I see, $u(v(x))$ would have been $x^{x^x}$ of course. That’s a bit embarrassing...
      – L. McDonald
      Nov 15 at 7:56
















    Sorry about the video, he did almost exactly the same thing that you did, but might I ask, would you be able to show me where I went wrong?
    – L. McDonald
    Nov 15 at 7:49




    Sorry about the video, he did almost exactly the same thing that you did, but might I ask, would you be able to show me where I went wrong?
    – L. McDonald
    Nov 15 at 7:49












    If $u(x)=x^x, v(x)=x^x$, then $u(v(x))=(x^x)^{x^x}=x^{x^{x+1}}$, which is a different function.
    – farruhota
    Nov 15 at 7:54




    If $u(x)=x^x, v(x)=x^x$, then $u(v(x))=(x^x)^{x^x}=x^{x^{x+1}}$, which is a different function.
    – farruhota
    Nov 15 at 7:54












    Oh, I see, $u(v(x))$ would have been $x^{x^x}$ of course. That’s a bit embarrassing...
    – L. McDonald
    Nov 15 at 7:56




    Oh, I see, $u(v(x))$ would have been $x^{x^x}$ of course. That’s a bit embarrassing...
    – L. McDonald
    Nov 15 at 7:56










    up vote
    2
    down vote













    For differentiating expressions of the form $u(x)^{v(x)},$ I've always found it easiest to take logarithms and then differentiate implicitly. However, rather than work with the more general form $y = u^v,$ I'll work with $y = x^U$ (i.e. use $x$ for the base), which is sufficient to deal with $x^{x^x}.$



    Taking the logarithm of both sides gives



    $$ y = x^U ;; implies ln y = lnleft(x^U right) ;; implies ;; ln y = U cdot ln x $$



    Now differentiate both sides with respect to $x$:



    $$ frac{d}{dx}left(ln yright) ;; = ;; frac{d}{dx}left(U cdot ln x right) $$



    $$ frac{y'}{y} ;; = ;; frac{d}{dx}(U) cdot ln x ; + ; U cdot frac{d}{dx} (ln x) ;; = ;; U'ln x ; + ; frac{U}{x} $$



    $$ y' ;; = ;; y cdot left[ U'ln x ; + ; frac{U}{x} right] ;; = ;; x^U cdot left[ U'ln x ; + ; frac{U}{x} right] $$



    Using this result for $U(x) = x,$ we get



    $$ left(x^xright)' ;; = ;; x^x cdot left[ 1 cdot ln x ; + ; frac{x}{x} right] ;; = ;; x^x(ln x + 1) $$



    Using that same result for $U(x) = x^x$ along with what we just found for $left(x^xright)',$ we get



    $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$



    $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$



    $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$



    $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$






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      For differentiating expressions of the form $u(x)^{v(x)},$ I've always found it easiest to take logarithms and then differentiate implicitly. However, rather than work with the more general form $y = u^v,$ I'll work with $y = x^U$ (i.e. use $x$ for the base), which is sufficient to deal with $x^{x^x}.$



      Taking the logarithm of both sides gives



      $$ y = x^U ;; implies ln y = lnleft(x^U right) ;; implies ;; ln y = U cdot ln x $$



      Now differentiate both sides with respect to $x$:



      $$ frac{d}{dx}left(ln yright) ;; = ;; frac{d}{dx}left(U cdot ln x right) $$



      $$ frac{y'}{y} ;; = ;; frac{d}{dx}(U) cdot ln x ; + ; U cdot frac{d}{dx} (ln x) ;; = ;; U'ln x ; + ; frac{U}{x} $$



      $$ y' ;; = ;; y cdot left[ U'ln x ; + ; frac{U}{x} right] ;; = ;; x^U cdot left[ U'ln x ; + ; frac{U}{x} right] $$



      Using this result for $U(x) = x,$ we get



      $$ left(x^xright)' ;; = ;; x^x cdot left[ 1 cdot ln x ; + ; frac{x}{x} right] ;; = ;; x^x(ln x + 1) $$



      Using that same result for $U(x) = x^x$ along with what we just found for $left(x^xright)',$ we get



      $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$



      $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$



      $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$



      $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        For differentiating expressions of the form $u(x)^{v(x)},$ I've always found it easiest to take logarithms and then differentiate implicitly. However, rather than work with the more general form $y = u^v,$ I'll work with $y = x^U$ (i.e. use $x$ for the base), which is sufficient to deal with $x^{x^x}.$



        Taking the logarithm of both sides gives



        $$ y = x^U ;; implies ln y = lnleft(x^U right) ;; implies ;; ln y = U cdot ln x $$



        Now differentiate both sides with respect to $x$:



        $$ frac{d}{dx}left(ln yright) ;; = ;; frac{d}{dx}left(U cdot ln x right) $$



        $$ frac{y'}{y} ;; = ;; frac{d}{dx}(U) cdot ln x ; + ; U cdot frac{d}{dx} (ln x) ;; = ;; U'ln x ; + ; frac{U}{x} $$



        $$ y' ;; = ;; y cdot left[ U'ln x ; + ; frac{U}{x} right] ;; = ;; x^U cdot left[ U'ln x ; + ; frac{U}{x} right] $$



        Using this result for $U(x) = x,$ we get



        $$ left(x^xright)' ;; = ;; x^x cdot left[ 1 cdot ln x ; + ; frac{x}{x} right] ;; = ;; x^x(ln x + 1) $$



        Using that same result for $U(x) = x^x$ along with what we just found for $left(x^xright)',$ we get



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$






        share|cite|improve this answer














        For differentiating expressions of the form $u(x)^{v(x)},$ I've always found it easiest to take logarithms and then differentiate implicitly. However, rather than work with the more general form $y = u^v,$ I'll work with $y = x^U$ (i.e. use $x$ for the base), which is sufficient to deal with $x^{x^x}.$



        Taking the logarithm of both sides gives



        $$ y = x^U ;; implies ln y = lnleft(x^U right) ;; implies ;; ln y = U cdot ln x $$



        Now differentiate both sides with respect to $x$:



        $$ frac{d}{dx}left(ln yright) ;; = ;; frac{d}{dx}left(U cdot ln x right) $$



        $$ frac{y'}{y} ;; = ;; frac{d}{dx}(U) cdot ln x ; + ; U cdot frac{d}{dx} (ln x) ;; = ;; U'ln x ; + ; frac{U}{x} $$



        $$ y' ;; = ;; y cdot left[ U'ln x ; + ; frac{U}{x} right] ;; = ;; x^U cdot left[ U'ln x ; + ; frac{U}{x} right] $$



        Using this result for $U(x) = x,$ we get



        $$ left(x^xright)' ;; = ;; x^x cdot left[ 1 cdot ln x ; + ; frac{x}{x} right] ;; = ;; x^x(ln x + 1) $$



        Using that same result for $U(x) = x^x$ along with what we just found for $left(x^xright)',$ we get



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$



        $$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$







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        edited Nov 15 at 7:09

























        answered Nov 15 at 6:43









        Dave L. Renfro

        24k33979




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