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.
calculus derivatives chain-rule tetration
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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.
calculus derivatives chain-rule tetration
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
add a comment |
up vote
0
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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.
calculus derivatives chain-rule tetration
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
calculus derivatives chain-rule tetration
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
add a comment |
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
add a comment |
2 Answers
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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).
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
add a comment |
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
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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).
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
add a comment |
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).
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
add a comment |
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).
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).
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
add a comment |
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
add a comment |
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
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$
add a comment |
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
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$
add a comment |
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
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$
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
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$
$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$
edited Nov 15 at 7:09
answered Nov 15 at 6:43
Dave L. Renfro
24k33979
24k33979
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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