Derivative of x^x^x - Logarithmic Differentiation of Exponential Functions

This calculus video explains how to find the derivative of x^x^x using logarithmic differentiation which is useful for differentiating exponential functions with a variable in the exponent position.

Derivatives - The Product Rule - f*g:

Derivatives - The Quotient Rule:

Derivatives - The Chain Rule:

Derivatives - Composite Functions:

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Implicit Differentiation:

Derivatives of Inverse Trig Functions:

Derivatives of Exponential Functions:

Derivatives of Logarithmic Functions:

Logarithmic Differentiation:

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Derivatives - Using Logarithms:

Derivatives of Inverse Functions:

Derivatives - Differentiation Rules:

Derivatives - Function Notations:

Derivatives - The Reciprocal Rule:

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Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

Derivatives - Formula Sheet:

[spondulix]

*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.

[TedHopp]

You have to be careful about the domain. In particular, ln(ln(x)) is not defined for x<=1.

[zenandelusizi]

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)

[aimanali]

ma sha allah ma sha allah what a tutor ma sha allah god bless you brother

[lambthor]

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 ;)

[shielamariekadusale]

Can you solve y= log ³√√12x in derivative of differentiation Logarithmic Functions?

[priyasmitasahu]

Thank You for the video... I can understand easily... Thanks a lot ❤️❤️❤️

[MrJatinSingh]

please upload a video about Limits, Continuity and differentiability.

[MisutaaPurinsu]

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]]'

[kevinantiporta]

Lumabas to sa search for mathematician of the year sa amin sa school

[usamanavid]

How to solve integration of
(a^x^x^x).(a^x^x).(a^x)

[ribkamoses]

(a^m)^n=a^mn why we should not use that here

[SierraHunter]

Wont the two x on the top multiply out to give x^(x^2) ?? Exponent rule

[spondulix]

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.

[thamsanqakendriege]

I think if you let u = x^x its much simpler

[newton]

wouldn't chain rule be easier? it gives 2X^(X^x)^2 * 2X^X^2 * 2X^2

[longingsummers]

You complicated it even more, why can't we use derivative of product rule

[ModerniteGamer-vfrh]

So lengthy solution. It was a simple question you complicated it even more . It could have been solved by simpler method 🥴🥴

[vistalabel]

4:01 u' why it is 9x? instead of 9

[darcash]

anyone else just do derivative x^x^x^x^x after this just for the hell of it? 🥴