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*LEANING TOWER OF EXPONENTS (LTE’s)* : This video purports to find the derivative with respect to x of the leaning tower of identical exponents x of order three. However, it actually finds the derivative with respect to x of x to the LTE of order two. Except at the point x = 2, these are not equivalent. In the context of ordinary numbers, a number or function raised to a power is annotated universally by placing the power above and to the right of the number or function. For example, the b-th power of a is written with b above and to the right of a. In like manner, the c-th power of the b-th power of a is written with c placed above and to the right of the notation for the b-th power of a. This gives rise to an open (no parentheses) leaning tower, or LTE, consisting of the sequence a, b and c. In in-line notation, this is written as a^b^c, where the circumflex ^ represents exponentiation. Parentheses can be added to make the implied order of exponentiation explicit. Viz, a^b^c = ((a^b)^c) and a^b^c^d = (((a^b)^c)^d). So, if a, b, c and d all equal x, then x^x^x = ((x^x)^x) and x^x^x^x = (((x^x)^x)^x). With a, b and c all equal to x, this video finds the derivative with respect to x of a^b^c = (a^(b^c)) = x^x^x = (x^(x^x)) and not of a^b^c = x^x^x = ((a^b)^c) = ((x^x)^x). In the former case, ln (x^(x^x)) = (x^x) ln x, whereas, in the latter case, ln ((x^x)^x) = x ln (x^x). The former follows from the non-standard interpretation assumed in the video and the latter follows from the standard interpretation of exponentiation. See equation ln y = ln (x^(x^x)) = (x^x) ln x, the equation completed at 1:13 into the video. When a, b, and c, etc., are all equal to, say, x, then x^x^x is referred to as the third tetration of x. Likewise, x^x^x^x is referred to as the fourth tetration of x. I suggest the in-line notations ^3x, ^4x and ^nx for the third, fourth and n-th tetrations of x, respectively, though parentheses may be required to disambiguate the application of the specified exponent to the right, as in these examples, from application of the exponent to the left as in the application of the exponent ^b to the left in a^b = "a to the power b". Consistent with the interpretation of x^x^x employed in the video, x^x^x^x is interpreted as x^(x^x^x), and, consequently, ln x^x^x^x equals (x^x^x) ln x, where the factor preceding ln x comes from the final three factors of x^x^x^x and not from the first three factors of x^x^x^x. Since, from the first sentence in the second previous paragraph above, x^x^x = x^(x^x), substitution yields x^x^x^x = x^(x^x^x) = x^(x^(x^x)) in the interpretation of the video, which interpretation is not consistent with the universally standard way of representing and interpreting exponentiation of exponentiations. Note from the fourth previous paragraph that the standard interpretation of exponentiation expressed in in-line notation reads x^x^x^x = (((x^x)^x)^x), whereas in the non-standard interpretation used in the video x^x^x^x = (x^(x^(x^x))). One sees that the pattern of nested parentheses proceeds from the left and outward in the standard interpretation of exponentiation, and from the right-and outward in the non-standard interpretation employed by the video. These patterns extend in the obvious way to LTE’s of higher order and, of course, to tetrations of higher order. When the proposed tetration notation is used to express and compare the standard and non-standard interpretations of exponentiation, one finds, for example, that x^x^x = ^3x = (^2x)^x (standard) x^x^x = ^3x = x^(^2x) (non-standard) x^x^x^x = ^4x = ((^2x)^x)^x (standard) x^x^x^x = ^4x = x^(x^(^2x)) (non-standard) x^x^x^x^x = ^5x = (((^2x)^x)^x)^x standard) x^x^x^x^x = ^5x = x^(x^(x^(^2x))) (non-standard) The pattern trends evident above extend in the obvious way to tetrations of arbitrary integer order. One sees that the in-line representations of the two interpretations are mirrors of each other. For any positive integer n, the two interpretations of the tetration ^nx are inequivalent except when x = 2.
Thanks sir , this videos very much helpful.The steps are a little bit similar to y=x^ x, i would have gotten the answer if i new you have to take the natural log twice (wink)
Could have just used the formula for y= f^g (f & g two variables) right? i.e y' = f^g[g'Lnf + g(f'/f)].. Solved it that way in 30 secs... bt never a bad idea to explore ;)
a little late maybe. But its because a^m^n is not equal to (a^m)^n, its equal to a^(m^n) 2^3^4 = 2.4178516392 * 10^24 2^(3^4) = 2.4178516392 * 10^24 (2^3)^4 = 4096 2^(3*4) = 4096
y=x^x^x^x Apply ln throughout lny=x^x^xlnx Apply ln again throughout lnlny=ln(x^x^xlnx) lnlny=x^xlnx+lnlnx You need to apply ln again lnlnlny=ln(x^xlnx+lnlnx), here its very complicated one since ln can't get in the bracket due to the presence of "+". For me I can't go any further
@@darcash1738 No, I am not joking. I believe that the derivation in this TH-cam clip is quite incorrect. It must contain a subtle error somewhere, though I haven’t spotted it yet myself. However, it is easy to verify that the TH-cam formula is incorrect. Here is the derivation of my result, which I have confirmed yields correct results. 1. y = ((x^x)^x) 2. ln y = x ln (x^x) = x^2 ln x 3. d ln y/dx = 1/y (dy/dx) or dy/dx = y (d ln y/dx) From 2, 4. d ln y /dx = 2 x ln x + x^2 (d ln x/dx) = 2 x ln x + x = x (2 ln x + 1) Finally, 1 and 4 in 3 yields, 5. dy/dx = ((x^x)^x) [x (2 ln x +1) ] BTW, in my posting I wrote (x^x^x) where I should have written ((x^x)^x). I’ll correct that shortly. If you evaluate 5 above for x = 1.5 and x = 1.8, you get the following values for dy/dx. dy/dx = 6.673917582 for x = 1.5 dy/dx = 26.298389167 for x = 1.8 If you have a calculator that will evaluate any number to any power you can estimate dy/dx numerically. For example, compute ((x^x)^x) for x = 1.50001 and again for x = 1.50000 and form the difference divided by the interval 0.00001. You will see that the numerical result is very close to the result above from my formula for x = 1.5. Similarly for x = 1.8. On the other hand, the formula derived in the TH-cam clip yields erroneous values for these, and other, test values. BTW, you state as evidence that his result must be correct that "he is the organic chem tutor after all lmao". FWIW, I have a Ph.D. in theoretical physics. That is, mathematical physics. Please let me know if you have any questions about my derivation.
@@darcash1738 BTW, if you followed my previous derivation, you should be able to follow the derivation below for x to a totem pole of n powers of x. 1. y = ((((x^x)^x)^x) . . . . ) for n x’s. Repeated application of natural log yields 2. ln y = = x^(n - 1) ln x 3. d ln y/dx = 1/y (dy/dx) or dy/dx = y (d ln y/dx) From 2, 4. d ln y /dx = (n - 1) x^(n - 2) ln x + x^(n - 2) = x^(n - 2) [ (n - 1) ln x + 1] Finally, 1 and 4 in 3 yields, 5. dy/dx = ((((x^x)^x)^x) . . . . ) x^(n - 2) [ (n - 1) ln x +1) ]
@@spondulix99 it took me a minute, but I’ve located where you got thrown off track. (x^x)^x is not the same as x^x^x. With the first one, you can multiply out the exponents, resulting in x^(x^2). Because of that, step 2 isn’t right. This step here was critical to the application of the log identity he uses later-ln[x^(x^x)], where you move the power out front-making it x^x ln(x) and not x^2 ln(x)
@@darcash1738 I have identified the error in this TH-cam derivation of the derivative of ((x^x)^x). Right at the beginning, he states incorrectly that 1. ln y = ln ((x^x)^x) = (x^x) ln x. This is not correct and causes his final result to be incorrect. The correct relationship is, 2. ln y = ln ((x^x)^x) = x ln (x^x) = x^2 ln x. For example, assume that x = 1.5. Then 3. ((x^x)^x) = 2.490034319 so that ln y = ln ((x^x)^x) = 0.912296493 Likewise, correct equation 2 gives 4. ln y = x^2 ln x = 2.25 ln (1.5) = 0.912296493 But, incorrect equation 1 gives 5. ln y = (x^x) ln x = 1.837117307 ln (1.5) = 0.744886968 Only when x = 2 is 1 correct.
Yes. I have obtained exactly that result. In fact, here's what I got: [x^x^g(x)]' = [x^x^g(x)] [ g'(x) x ln x + g(x) ( ln x + 1) ] where prime means d/dx and x is real positive. In the special case of g(x) = 1 one gets the result for the derivative of x^x, and if g(x) = x one gets your result for the derivative of x^x^x.
What will be the general formula for the derivative of x^x^x^x ..... upto N times where N is any integer ( not an infinity ) ? If you will tell, then I will consider you as an intelligent person ....... perhaps even now you may be intelligent .........
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Derivatives - Free Formula Sheet: bit.ly/4dThzf1
Final Exams and Video Playlists: www.video-tutor.net/
Full-Length Math & Science Videos: www.patreon.com/mathsciencetutor/collections
*LEANING TOWER OF EXPONENTS (LTE’s)* : This video purports to find the derivative with respect to x of the leaning tower of identical exponents x of order three. However, it actually finds the derivative with respect to x of x to the LTE of order two. Except at the point x = 2, these are not equivalent.
In the context of ordinary numbers, a number or function raised to a power is annotated universally by placing the power above and to the right of the number or function.
For example, the b-th power of a is written with b above and to the right of a. In like manner, the c-th power of the b-th power of a is written with c placed above and to the right of the notation for the b-th power of a. This gives rise to an open (no parentheses) leaning tower, or LTE, consisting of the sequence a, b and c. In in-line notation, this is written as a^b^c, where the circumflex ^ represents exponentiation.
Parentheses can be added to make the implied order of exponentiation explicit. Viz, a^b^c = ((a^b)^c) and a^b^c^d = (((a^b)^c)^d). So, if a, b, c and d all equal x, then x^x^x = ((x^x)^x) and x^x^x^x = (((x^x)^x)^x).
With a, b and c all equal to x, this video finds the derivative with respect to x of a^b^c = (a^(b^c)) = x^x^x = (x^(x^x)) and not of a^b^c = x^x^x = ((a^b)^c) = ((x^x)^x). In the former case, ln (x^(x^x)) = (x^x) ln x, whereas, in the latter case, ln ((x^x)^x) = x ln (x^x). The former follows from the non-standard interpretation assumed in the video and the latter follows from the standard interpretation of exponentiation. See equation ln y = ln (x^(x^x)) = (x^x) ln x, the equation completed at 1:13 into the video.
When a, b, and c, etc., are all equal to, say, x, then x^x^x is referred to as the third tetration of x. Likewise, x^x^x^x is referred to as the fourth tetration of x. I suggest the in-line notations ^3x, ^4x and ^nx for the third, fourth and n-th tetrations of x, respectively, though parentheses may be required to disambiguate the application of the specified exponent to the right, as in these examples, from application of the exponent to the left as in the application of the exponent ^b to the left in a^b = "a to the power b".
Consistent with the interpretation of x^x^x employed in the video, x^x^x^x is interpreted as x^(x^x^x), and, consequently, ln x^x^x^x equals (x^x^x) ln x, where the factor preceding ln x comes from the final three factors of x^x^x^x and not from the first three factors of x^x^x^x. Since, from the first sentence in the second previous paragraph above, x^x^x = x^(x^x), substitution yields x^x^x^x = x^(x^x^x) = x^(x^(x^x)) in the interpretation of the video, which interpretation is not consistent with the universally standard way of representing and interpreting exponentiation of exponentiations.
Note from the fourth previous paragraph that the standard interpretation of exponentiation expressed in in-line notation reads x^x^x^x = (((x^x)^x)^x), whereas in the non-standard interpretation used in the video x^x^x^x = (x^(x^(x^x))). One sees that the pattern of nested parentheses proceeds from the left and outward in the standard interpretation of exponentiation, and from the right-and outward in the non-standard interpretation employed by the video. These patterns extend in the obvious way to LTE’s of higher order and, of course, to tetrations of higher order.
When the proposed tetration notation is used to express and compare the standard and non-standard interpretations of exponentiation, one finds, for example, that
x^x^x = ^3x = (^2x)^x (standard)
x^x^x = ^3x = x^(^2x) (non-standard)
x^x^x^x = ^4x = ((^2x)^x)^x (standard)
x^x^x^x = ^4x = x^(x^(^2x)) (non-standard)
x^x^x^x^x = ^5x = (((^2x)^x)^x)^x standard)
x^x^x^x^x = ^5x = x^(x^(x^(^2x))) (non-standard)
The pattern trends evident above extend in the obvious way to tetrations of arbitrary integer order.
One sees that the in-line representations of the two interpretations are mirrors of each other. For any positive integer n, the two interpretations of the tetration ^nx are inequivalent except when x = 2.
Great comment, shame that yt comments doon't have latex or something for pretty equation formatting
You have to be careful about the domain. In particular, ln(ln(x)) is not defined for x
Thanks sir , this videos very much helpful.The steps are a little bit similar to y=x^ x, i would have gotten the answer if i new you have to take the natural log twice (wink)
It is xˣ•(ln(x)+1)
I prefer benzene reactions.
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does benzene give cancer?
😂
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I am from India 🇮🇳
Proud to b an Indian
ma sha allah ma sha allah what a tutor ma sha allah god bless you brother
Could have just used the formula for y= f^g (f & g two variables) right? i.e y' = f^g[g'Lnf + g(f'/f)].. Solved it that way in 30 secs... bt never a bad idea to explore ;)
Thank You for the video... I can understand easily... Thanks a lot ❤️❤️❤️
easily you say?
Calc exam on Tuesday
I want to cry
how was ıt
Can you solve y= log ³√√12x in derivative of differentiation Logarithmic Functions?
لا إله إلا الله
please upload a video about Limits,Continuity and differentiability.
Here you go:
th-cam.com/video/6e4Wtgc43KQ/w-d-xo.html
Another method is switching x^x to e^ln(x^x) supposedly that we know how to derivate e^u {[e^u]'=u'e^u}
You end up with [e^[lnx * e^xlnx]]'
Lumabas to sa search for mathematician of the year sa amin sa school
Very nicely explained
Interesting question, thanks a lot!
Quite the philosopher you are
anyone else just do derivative x^x^x^x^x after this just for the hell of it? 🥴
How to solve integration of
(a^x^x^x).(a^x^x).(a^x)
Same procedure
@@kidkesReacts haha, about 4 years too late
Great video!
Never misses😭🤙🏽🔥
Oh man you saved me❤️❤️❤️
Thanks for this derivatives
OCT does it again!
Very good
(a^m)^n=a^mn why we should not use that here
Ikr
a little late maybe. But its because a^m^n is not equal to (a^m)^n, its equal to a^(m^n)
2^3^4 = 2.4178516392 * 10^24
2^(3^4) = 2.4178516392 * 10^24
(2^3)^4 = 4096
2^(3*4) = 4096
Thanks a lot sir
Wont the two x on the top multiply out to give x^(x^2) ?? Exponent rule
How to solve x^x^x^x = y. Dy/dx?
Complicated one!!!
y=x^x^x^x
Apply ln throughout
lny=x^x^xlnx
Apply ln again throughout
lnlny=ln(x^x^xlnx)
lnlny=x^xlnx+lnlnx
You need to apply ln again
lnlnlny=ln(x^xlnx+lnlnx), here its very complicated one since ln can't get in the bracket due to the presence of "+". For me I can't go any further
I think if you let u = x^x its much simpler
Thank u very much
👍useful
wouldn't chain rule be easier? it gives 2X^(X^x)^2 * 2X^X^2 * 2X^2
I never seen it used that way. I'm surprised there are no logs or natural logs which is usually present when differentiating exponential functions.
it's been a while since calc but thats how I think I would have done it. (dx) X^u where u = X^x is (X^u)' * u' where u' = (X^v)' * v' and v=x
I'm certainly not in lowest terms though.
Are you certain that your final result is correct? I get: d ((x^x)^x) / dx = ((x^x)^x) [ x (2 ln x + 1) ] for x real positive.
I did it without watching him and got the same result, so unless youre joking, it's correct. he is the organic chem tutor after all lmao
@@darcash1738 No, I am not joking. I believe that the derivation in this TH-cam clip is quite incorrect. It must contain a subtle error somewhere, though I haven’t spotted it yet myself. However, it is easy to verify that the TH-cam formula is incorrect. Here is the derivation of my result, which I have confirmed yields correct results.
1. y = ((x^x)^x)
2. ln y = x ln (x^x) = x^2 ln x
3. d ln y/dx = 1/y (dy/dx) or dy/dx = y (d ln y/dx)
From 2,
4. d ln y /dx = 2 x ln x + x^2 (d ln x/dx) = 2 x ln x + x = x (2 ln x + 1)
Finally, 1 and 4 in 3 yields,
5. dy/dx = ((x^x)^x) [x (2 ln x +1) ]
BTW, in my posting I wrote (x^x^x) where I should have written ((x^x)^x). I’ll correct that shortly.
If you evaluate 5 above for x = 1.5 and x = 1.8, you get the following values for dy/dx.
dy/dx = 6.673917582 for x = 1.5
dy/dx = 26.298389167 for x = 1.8
If you have a calculator that will evaluate any number to any power you can estimate dy/dx numerically. For example, compute ((x^x)^x) for x = 1.50001 and again for x = 1.50000 and form the difference divided by the interval 0.00001. You will see that the numerical result is very close to the result above from my formula for x = 1.5. Similarly for x = 1.8.
On the other hand, the formula derived in the TH-cam clip yields erroneous values for these, and other, test values.
BTW, you state as evidence that his result must be correct that "he is the organic chem tutor after all lmao". FWIW, I have a Ph.D. in theoretical physics. That is, mathematical physics.
Please let me know if you have any questions about my derivation.
@@darcash1738 BTW, if you followed my previous derivation, you should be able to follow the derivation below for x to a totem pole of n powers of x.
1. y = ((((x^x)^x)^x) . . . . ) for n x’s.
Repeated application of natural log yields
2. ln y = = x^(n - 1) ln x
3. d ln y/dx = 1/y (dy/dx) or dy/dx = y (d ln y/dx)
From 2,
4. d ln y /dx = (n - 1) x^(n - 2) ln x + x^(n - 2) = x^(n - 2) [ (n - 1) ln x + 1]
Finally, 1 and 4 in 3 yields,
5. dy/dx = ((((x^x)^x)^x) . . . . ) x^(n - 2) [ (n - 1) ln x +1) ]
@@spondulix99 it took me a minute, but I’ve located where you got thrown off track. (x^x)^x is not the same as x^x^x.
With the first one, you can multiply out the exponents, resulting in x^(x^2). Because of that, step 2 isn’t right.
This step here was critical to the application of the log identity he uses later-ln[x^(x^x)], where you move the power out front-making it x^x ln(x) and not x^2 ln(x)
@@darcash1738 I have identified the error in this TH-cam derivation of the derivative of ((x^x)^x).
Right at the beginning, he states incorrectly that
1. ln y = ln ((x^x)^x) = (x^x) ln x.
This is not correct and causes his final result to be incorrect. The correct relationship is,
2. ln y = ln ((x^x)^x) = x ln (x^x) = x^2 ln x.
For example, assume that x = 1.5. Then
3. ((x^x)^x) = 2.490034319 so that ln y = ln ((x^x)^x) = 0.912296493
Likewise, correct equation 2 gives
4. ln y = x^2 ln x = 2.25 ln (1.5) = 0.912296493
But, incorrect equation 1 gives
5. ln y = (x^x) ln x = 1.837117307 ln (1.5) = 0.744886968
Only when x = 2 is 1 correct.
You complicated it even more, why can't we use derivative of product rule
So lengthy solution. It was a simple question you complicated it even more . It could have been solved by simpler method 🥴🥴
Thanks a lot
it's best method
how about y=x^x^y
Thank you sir
Thank you
why do you take the natural log for the second time?
Taking the ln a second time will let you apply the logarithmic exponent property a second time. Then you will have no more variable exponents.
Не нашёл решения на русских сайтах , решил зайти на англоязычные и о чудо
Rd Sharma book me hai ye as example
4:01 u' why it is 9x? instead of 9
might be a mistake
4:09 i know 4 years too late. But literally could’ve watched 8 seconds more before asking.
Who had the answer coming x^x^x[2xlogx+1] .....anyone?
Yes. I have obtained exactly that result. In fact, here's what I got: [x^x^g(x)]' = [x^x^g(x)] [ g'(x) x ln x + g(x) ( ln x + 1) ] where prime means d/dx and x is real positive. In the special case of g(x) = 1 one gets the result for the derivative of x^x, and if g(x) = x one gets your result for the derivative of x^x^x.
Love you
How about x^y^x^y^x^y dx/dy?
you da mon
👍
ıt was tiring...
Wow
Op
1rd
😂
What will be the general formula for the derivative of x^x^x^x ..... upto N times where N is any integer ( not an infinity ) ?
If you will tell, then I will consider you as an intelligent person ....... perhaps even now you may be intelligent .........
Thank you