Could you maybe explain why we cannot take d/dx(x^x)? YouTube viewer problem

[@Steve_Stowers]

As usual, the video gives a good, detailed, accurate explanation. But the short answer is: when applying derivative rules, you have to pay attention to what's a constant and what's a variable.

[@MikehMike01]

The ‘dx’ is not decoration, it tells you everything

[@knivetsil]

That satisfied smirk at saving the almost-miswritten “x” in the exponent at 8:55 😂😂

[@imaginaryangle]

I love how you never forget what it's like to look at a problem without all the abstract knowledge that comes on top of it later, and are able to explain every concept at the appropriate level of intuition. You are a wonderful teacher and your videos are always fun and full of good vibes!

[@theupson]

your method is absolutely the industry standard, but the multivariate chain rule is so much cleaner conceptually that i will generally take this error as an opportunity to teach it even to a first semester calculus student.

[@Jesin00]

Which formulation do you use? I find the version in my textbook has too many pieces for many students to keep track of.

[@db5094]

I would also like to know​@@Jesin00

[@djapa9453]

It has always been interesting to me that the derivative of x^x is equal to x^x+lnx * x^x which is what you get if you first treat the x in the exponent as a constant added to if you treat the base as a constant

[@person1082]

multivariable chain rule

[@tobybartels8426]

That's not a coincidence!

[@Axacqk]

@@person1082 PLUS the idea to treat the problem as multivariate.

[@person1082]

you can also use multivariable chain rule: d/dx(x^x)=x x^(x-1)+x^x ln(x)

[@APaleDot]

Yo, that's crazy.

[@-Hellow-]

could u explain?

[@APaleDot]

@@-Hellow- If you have a function of two variables f(x, y) then the total derivative of that function is given by the multivariable chain rule: df/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt) Applying this the function f(x, y) = x ^ y gives you the following equation because ∂f/∂x treats 'y' as a constant and ∂f/∂y treats 'x' as a constant : df/dt = (y · x ^ (y - 1))(dx/dt) + (ln(x) · x ^ y)(dy/dt) Now, since we're actually interested in x ^ x, we just set y = t = x, which makes dx/dt = dy/dt = dx/dx = 1, leaving us with: df/dx = x · x ^ (x - 1) + ln(x) · x ^ x

[@cdkw8254]

I always thought implicit differenciation is used to take derivative of x^x but this method is better.

[@timecrystal5thapostleof125]

It's essentially the same thing since taking logs on both sides and doing e^ln(x) does the same thing. But I do agree that this approach feels more intuitive

[@cdkw8254]

Exactly@@timecrystal5thapostleof125

[@nathanbarnes3969]

I actually think the implicit differentiation method is more intuitive. Deriving a term with a y-variable, with respect to x, is basically just chain rule, it’s much faster than having to manipulate or re-write an equation in order to differentiate explicitly

[@mudi2000a]

Don’t know why I watched this because I already knew it but I find it incredible that you can explain it in less than 10 minutes including all the pitfalls.

[@nikitakipriyanov7260]

When I was 15, I was tasked to find this derivative. I've invented a way to solve this (and solved correctly) in a less than a minute. The answer is (x^x)' = (ln x + 1) * x^x. My way was: I noticed that (ln f(x))' = f'(x)/f(x). Then, f'(x) = f(x) * (ln f(x))', so if f(x) = x^x, then (x^x)' = x^x * (ln(x^x))' = x^x * (x ln x)' = x^x * (ln x + 1)

[@Axacqk]

About that same age I figured it out by imagining f(x,y)=x^y as a surface, f(x,x)=x^x as its intersection with the y=x plane, the plane tangent to the surface at a point on the curve, the dx-sized parallelogram on said plane, etc. I only learned years later that this was basically the multivariate chain rule, except this way of thinking about it avoids certain complications of the full rule that later cancel out anyway.

[@WhiteGandalfs]

Actually, the application of the power rule and the exponential rule is perfectly correct. As with all combined functions, you get the overall derivative by summing up the derivatives of all parts of variances independently from each other. For x^x: d(x^x) / dx = d(b^x) / dx + d(x^k) / dx - with b and k treated as constants, but having the value x result: 1. for the power derivative: k * x^(k-1) == x^k == x^x (remember: k==x) 2. for the exponential derivative: ln(b) * b^x == ln(x) * x^x (remember: b==x) sum: x^x * (1 + ln(x)) - identical to what was shown in the video. As always in the science of math: "All roads lead to Rome" - as long as you obey the rules.

[@iabervon]

I kinda wanted you to correct d/dx(3³) by pointing out that you need the chain rule, so it's 3(3²)d/dx(3). You can use the power rule on any expression with a constant exponent, but you need the chain rule, which usually means you would do better to simplify it first.

[@jagapata6073]

U can also do this if u take the x^x=y Then taking log on both side and derivative using product rule.

[@alanclarke4646]

An alternative method is to let y = x^x. Then take natural logs of both sides. Now differentiate both sides with respect to x. LHS will be (1/y) times (dy/dx). So multiply both sides by (1/y ) and simplify.

[@kavinesh_the_legend]

Thanks bprp

[@abgvedr]

8:57 Nice save 🙂

[@spenzr6920]

8:55*:)

[@NintendoGamer789]

It can be done directly from the limit definition, but it takes much longer and manipulations are not as intuitive

[@megabotvideos]

the graphs helped a lot with visualizing the derivatives

[@snared_]

Hey, thanks, this was a great video! Love that you included the d/dx (3^3) = -3*3^2- nonexample too. edit: was trying to strike through it like you have in the video but I guess youtube has changed the formatting. Sad.

[@567secret]

A way to see why you can't do it the same as the others is by the definition of the derivative. It's somewhat unhelpful when trying to find the result, but it will show you why it doesn't work.

[@erinsgeography3619]

I remember I saw a bad maths video explaining the derivative to be like this: Using the power rule, we get that it's (x)x^(x-1), which is the same as x^x Using the derivative of exponential, we get that it's x^x ln x Maybe we can just add them because these two methods seem fine, so x^x + x^x ln x. Believe it or not, that's the derivative of x^x

[@henryptung]

It's not quite just "add them because both methods seem fine". There's a stronger statement based on treating the expression as a two variable function f(u,v) = u^v, considering where this is differentiable (in a multivariate sense), and then computing d/dx f(x,x) using partial derivatives.

[@matchamitminze]

@@henryptung Yes, they know that, but they were referencing a type of joke video that shows that you can arrive at the correct answer using illegal math.

[@henryptung]

​@@matchamitminze Aye; I just think the _legal_ math backing the result is more interesting, and want people to know where this sum-of-derivatives pattern comes from (nice generalization of the product rule, really).

[@matchamitminze]

@@henryptung That’s fair. The total differential makes a lot of the “difficult” derivatives from calculus I quite trivial, which could usually be found if you used the limit definition of a derivative (if you wanted to do all the necessary work, lol).

[@Clawdragoons]

Interestingly, what you can do instead of using the e^ln() trick, is to just first treat the function as an exponential, then treat it as a power function. So if you're trying to take the derivative of some function, f^g, treating it as an exponential, you get: f^g*ln(f)*d/dx(g) With the d/dx(g) coming from the chain rule, since the exponent is some function of x. Then, treating it as a power rule, you get: g*f^(g-1)*d/dx(f) Again, the d/dx(f) coming from the chain rule, as the base g is a function of x. Now, just add them together: d/dx(f^g) = f^g*ln(f)*d/dx(g) + g*f^(g-1)*d/dx(f) This is simultaneously way easier, and also never taught, which is a weird combo. For the record, not only is the end result demonstrably correct, the logic behind it is, I think, sound. You're effectively just using partials.

[@nikitakipriyanov7260]

True!

[@craftcrewtv8094]

I was hoping for him to show the shortcut for this function. First pretend that the exponent x is a number and use the power rule, so we get x*x^(x-1) , then pretend that the base x is the number and do the other "rule", so we get x^x*(lnx) , then we simplify the first result to just x^x, factor it out and get that d/dx(x^x)=x^x(1+lnx) which is the correct answer.

[@tobybartels8426]

You can save the last wrong one using the Generalized Power Rule aka the Power Chain Rule. As long as the exponent is constant, the derivative of uⁿ is nuⁿ⁻¹ times the derivative of u. That applies when u is the independent variable (recovering the ordinary Power Rule), and it also applies (although it's overkill) when u is constant.

[@ianfowler9340]

Personally, I would explain 2^x and x^x, using logarithmic differentiation and avoid the x = e^(lnx) thing. For a novice, in my opinion, e^(lnx) adds more "magic" to the process unless you have clearly explained, in detail with practice, why a^[ log(base a) (y) ] = y. There is more to it than just stroking out the e and the ln. When I was teaching I never used this. Log diff goes back to basics and re-enforces a great technique. One less formula to memorize. Teach the short cuts much later in their development..

[@stealthgamer4620]

yeah i was gonna comment smth similar, just use logarithmic differentiation. i always get confused when there is e^lnx involved especially in differentiation so logarithmic diff. is simpler for me.

[@gregebert5544]

So, do we need to place some restrictions on x ? For example, if x >0 there are no obvious problems with x^x. But is x^x well-defined for x

[@telanis9]

x^x is always an element of the complex numbers if x is an element of the real numbers. There's no special case for non-integers. For the same reason n^x and x^n are also (for x, n in the real numbers): the complex numbers are an algebraically closed field.

[@telanis9]

A random example: (-pi)^(-pi) = -0.024757 + 0.011801i

[@ganeshgunjal9073]

Let y=x^x, Take log of both sides, ln(y)=x.ln(x), Take derivative (1/y)(dy/dx)=x(1/x)+ln(x) dy/dx=x^x(1+ln(x))

[@craftcrewtv8094]

I can't help it, this guy reminds me of mathologer, a lot. I'm don't know why, maybe it's his face. Mathologer also has a video about juggling.

[@FabienCournoyer]

You can also just do d/dx(x^N) + d/dx(N^x) = Nx^(N-1) + N^x ln N = (substitute N for x) x^x + x^x ln x. Partial derivatives my friend!

[@theupson]

start from the definition of the derivative (once!); you get this result and some good insight into the chain rule to boot.

[@octa5265]

Yup, the power rule applies only for variable^ constant and a^x log a is used for constant^variable, but x^x is variable^variable, so we treat it both ways.

[@adityagidde1688]

let y= x^x taking log on both sides, log y = log x^x log y = x log x diff. both sides wrt x, we get 1/y . dy/dx = 1+logx dy/dx = y (1+ log x) dy/dx = x^x (1+ log x) just like that....

[@iota8732]

Lol the huge stock of pens in the background is so great

[@TheRealBroodax]

I love the look at the camera at 8:55 when he writes the "x" as two curves because he made a typo.

[@akira_21_]

My first thought was to use the definition of derivative and just take a limit. Idk if it will give any result, i'll try to do it later

[@adarshgopalakrishnan9766]

I tried and it works

[@looz727]

My thoughts too, all the power/chain rules do helps us to find answers quicker but we tends to forget the restrictions that come with them When in doubt, just use the definition and derive from it

[@cheezeckez6843]

I’m in precalc rn… and everything in this video went over my head….

[@VedanthB9]

Don't worry, just enjoy the ride!

[@JakubS]

y=x^x lny=ln(x^x) lny=xlnx now we implicitly differentiate dy/dx*1/y=x/x+lnx dy/dx= y(1+lnx) since we know y=x^x, we get dy/dx=x^x*(1+lnx)

[@user-rc6ob7cm1p]

If you take this derivative in "both wrong ways", power rule and exponential rule and, and add them you get the correct result. Always.

[@person1082]

that happens because that’s what you would do to do the multivariable chain rule

[@mathisnotforthefaintofheart]

My office has no windows. One reason is because I want to smash the windows anytime I see someone making these kind of mistakes I call "fumbles". It fumes me. So I got this windowless office. I also don't eat M&M's when grading. I could mis-swallow a candy with disastrous consequences. Actually grading takes a toll on my emotional human mind. I need stress pills. Same with mistakes like (A+B)²=A²+B²....

[@BigTonyPersonal]

Sanest maths teacher

[@nothanks39]

username checks out

[@moerkx1304]

Maybe your students are doing the computations over a field with characteristic 2

[@puskar2.014]

Hey bprp! Can you do a theoretical video of what happens when you take a derivitive with respect to a constant? like: d/d5(x^2+9x-16) for example

[@wepped482]

So why not do the same thing with the 2^x instead of pointing out that it just is c^x*ln(c)? d/dx(2^x)=d/dx(e^ln(2^x)=d/dx(e^(xln(2)))=e^(xln(2))*d/dx(xln(2))=2^x*ln2

[@GPLB]

That’s the way to do d/dx(2^x). I think for the sake of time for this video he skipped through that explanation.

[@bodesshorts8640]

Can you try x or any number tetrated to i, if you can.do i tetrated to i

[@lessonkoirala480]

Let y=x^x. ln(y)=ln(x^x) ln(y)=xlnx. Implicit differentiation (1/y)dy/dx=(x/x)+lnx. dy/dx=y(1+lnx). dy/dx=x^x(1+lnx)

[@fordfactor]

When I show this derivative I like to expand that last result it so it can be written as ln(x^x^x) + x^x The tower of x's looks cool.

[@romanbykov5922]

Chen Lu must be a very good person that helps all the time.

[@MyNameIsSalo]

interesting to note that any function that has x as the exponent, will have a derivative where the initial function is part of the derivative multiplied by some scaling factor.

[@Erekose2023]

Since I actually had time when watching, thought I'd try iot first. Took the natural log of both sides and went down that track,. Same answer Was quite chuffed that my brain cells have not atrophied as much as I feared

[@Alan-mq9om]

Is it possible to find the derivative of x^x by writing it as y=x^x and using natural log on both sides then using implicit differentiation?

[@CorrectHorseBatteryStaple472]

5:10 "It's not 9, okay, it's 27" I don't know why this cracked me up but it did

[@mohammedamr1898]

I found a general rule for the all functions like this (f(x)^g(x))

[@jomariraphaellmangahas1991]

Can you make a video in your main channel about a function raise to a function in derivative and going back to the original function using differential equation?

[@technicallightingfriend4247]

I am ask my math teacher First Differentiation : d/dx( sinx) ^2= 2sinx.cosx.1 Okay Here actually A formula use d/dx(x^n) =nx^(n-1) But In Integration process If I have a Integrand (sinx)^2 Why? We can't use (Integrand)^n= 1/(n+1)(Integral) ^(n+1) +c Then my math teacher told Only one word "Because of It function 's" But I can't understand😅😅 So( ⚫🔴 ✒) Can you Elaborate the these types of things

[@matchamitminze]

The “reverse power rule” for integration works out for something like integrating x^2 to x^3/3 + C because the derivative of x with respect to itself is constant. If you’re trying to integrate (sin x)^2 with respect to x, then when you take the derivative of sin(x) (e.g. via u-substitution), the derivative is a non-constant function of x, so you now have to rewrite dx in terms of this non-constant function. In your example, if you’re trying to integrate sin^2(x)dx, you’d let u = sin(x) => du = cos(x)dx => du/cos(x) = dx. Now you’re integrating (u^2du)/cos(x), which isn’t entirely defined in terms of u (unless you restrict the domain and let x = arcsin(u), in which case, dx = du/sqrt(1-u^2), and thus, cos(x) = sqrt(1-u^2)), in which case, you’re now integrating u^2du/sqrt(1-u^2). To solve, you’d just use a trig substitution to undo the u-substitution, and you’d be integrating sin^2(x)dx/cos(x), like you saw before. This cos(x) term didn’t appear in the original problem, which means we’ve changed the integral. All in all, if you’re trying to do u-substitution, you need the derivative of u to already be present in the integral.

[@RubyPiec]

just curious: since d/dx(x^2)=2x. What does this mean? Since the slope is 2x, I'd assume you'd add 2x to the value to get the next one, but x+2x=3x and 3x is definitely not x^2 unless x is 3

[@De2Venner]

Your assumption is wrong. There are many ways to interpret the slope though. It says that close to a value A, the best linear approximation is (x+A)^2 = A^2 + (2A)x, which is kinda what you wanted to do I guess. Since the slope depends on x, it only gives local information about your function.

[@jamesharmon4994]

Please clarify something for me: you explained that the derivative of e^x = e^x. If x can be any value, why isn't the derivative of e^BOX = e^BOX?

[@jonathanchen1834]

i mean, x is just a variable, so d/dBOX e^BOX is e^BOX, but if we are talking about d/dx e^BOX, then BOX is a constant, so e^BOX will also be a constant, and the derivative of a constant is always zero.

[@ianfowler9340]

We are actually still using the chain rule here - we still need to multiple by d(BOX)/dx . But since BOX = x then d(BOX)/dx = 1 and we just don't bother to write d(e^x)/dx = e^x*(dx/dx). But if BOX = x^2 then we need the factor of 2x to get 2x*e^x.

[@jimmy_4]

Considering that we are plotting on x and y axis, having "e raised to any value" and having "e raised to x as variable" are two different things. Former means e raised to a real number which will always be same/constant for changing values of x (or in other words for any values of x). Later plots the graph of how e^x changes with changing values of x.

[@jimmy_4]

Considering that we are plotting y=e^x on x and y axis, having "x as any value" and having "x as variable" are two different things. Former means e raised to a real number which will always be same/constant for changing values of x (or in other words for any values of x). Later plots the graph of how e^x changes with changing values of x.

[@APaleDot]

Chain rule always applies, so if you have d(e^x) the chain rule tells you it should be equal to (e^x)·dx. Dividing by dx on both sides gives you d(e^x)/dx = e^x.

[@frederf3227]

Is the power rule valid iff the X^n and n is not in any way a function of X?

[@Steve_Stowers]

The n can be any constant. If it's not a constant, but it is not in any way a function of x, then you're getting into functions of more than one variable, and partial derivatives and things like that, which are normally covered in Calculus III.

[@paulyberk]

I loved the joke about calculus teachers giving 0s to their students 😅

[@josepherhardt164]

The explanation(s) given here is that the power rule just doesn't work, whereas other methods DO work. I understand this. I guess I was hoping for a deeper insight as to why the one method is inherently wrong, rather than realizing it just doesn't work.

[@MegaFootDude]

What's a derivative?

[@cosmicpsyops4529]

The slope of the tangent line.

[@cosmicpsyops4529]

First derivative is slope. Second is concavity.

[@johnwick7175]

Hehe the math teacher in bprp was a little fiery in this video :D

[@markgraham2312]

Great!

[@TheNewbiedrummer]

I don’t know nothing about that man, but I can tell you that maths is hard 😊

[@ritardstrength5169]

His students are very fortunate.

[@cv21a]

Excelllent! Thank you.

[@DarkTouch]

what about negative X?

[@carultch]

The original function isn't continuous and for the most part, is undefined for -x, when limited to the real numbers. When you expand x^x to complex numbers, the function spirals around the real x-axis, and selectively lands on real numbers in the real x-y plane. This is also the derivative for the function when extended to all real number inputs, allowing complex outputs as well.

[@SeekingTheLoveThatGodMeans7648]

should we say ln|x| here?

[@stealthgamer4620]

i believe you don’t because x^x is ALWAYS positive and that function is faster than ln(x) no matter what. If you distribute you could rewrite it as x^x + x^x*ln(x) *Although x^x *ln(x) is greater than x^x, the ln(x) is still outrun by its multiplicative argument x^x, so it doesn’t matter.* OR another way to look at is if you factor out x^x, its is faster/greater than (1+lnx) and as stated before we know x^x is only in the domain of [0, infinity]. Very similar to sqrt(x), or well we could say they have the same domain. For more insight, refer to the graph of the functions x^x, its derivative, and sqrt(x). I don’t have an explanation for you though on why x^x domain is greater or equal to 0. But we do know if it was an even root, x can’t be negative or else it’s an imaginary number.

[@matchamitminze]

@@stealthgamer4620 If you define exponentiation to be a^b = e^(b*ln(a)), then the domain of x^x (when considering real numbers only) cannot contain 0, restricting it to (0, +inf). This is because the domain of ln(x) when considering only real numbers is defined over (0, +inf), excluding 0.

[@stealthgamer4620]

@@matchamitminze Thank you for the clarification. Yeah usually I don’t rewrite in terms of e and ln power, which is a fault of mine. And tbh I forget ln(0) is undefined.

[@jimmy_4]

Could you maybe explain why derivative of e^x is e^x?

[@doriansw305]

Because the ln(e) = 1. So if you apply the exponential rule of differentiation, you get d/dx (e^x) = e^x * ln(e) => just e^x

[@ianfowler9340]

A lot of teachers find the derivative of ln(x) first using the limit definition of the derivative and then use implicit differentiation to to get the derivative of e^x. Thus requires a ln/lim swap which most don't justify proerly. If you use the limit definition directly on e^x you have to deal with (e^h -1)/h as h ----> 0 = 1. Again, hard to justify at the novice level. It's always a chicken before the egg problem.

[@lessonkoirala480]

Use the implicit differentiation to prove so if y=e^x. Ln(y)=x ln(e). ln(y)=x. (1/y)dy/dx=1 dy/dx=y. dy/dx=e^x

[@ianfowler9340]

@@lessonkoirala480 Yes, but your method requires the derivative of the ln(y). How are you going to explain where that comes from? You can't use the x = e^y and use implicit diff. to get derivative ln(x) because you will be using the result you are trying to prove. It's just a circular argument. Chicken before the egg.

[@ianfowler9340]

@@doriansw305 You are assuming the result you are trying to prove - i.e. you are using the exponential rule of differentiation to prove exponential rule of differentiation. Sorry, but that's a no-go.

[@theeraphatsunthornwit6266]

this video reminds me why I don't like math much in high school

[@saharhaimyaccov4977]

1:50 but if u use derrevitive by d/d3 (3³) this just like first question😂

[@Neodynium.the_permanent_magnet]

At least, could say d/dx(3^3) = d/dx(3 x 3^2) 😁

[@LucenProject]

Yay! Now X to the X to the X

[@Michael-sb8jf]

been a while since I done a problem like this I just figured you use the exponent rules and you get f'(x)=lnx x^x boy was I wrong

[@xinpingdonohoe3978]

It's strange. You use exponent rules, and you use the power rule, and you add them together. That's not a rigourous definition, but true in this case. (x^x)' = x^x ln(x) + x x^(x-1)

[@kicorse]

@@xinpingdonohoe3978 That's right, and it's not a coincidence. It results from the linearity of calculus. The two common mistakes are to treat the x in the exponent as a constant (which would give x^x), or to treat the x in the base as a constant (which would give ln(x) x^x). But analogous to the product rule (for example), it is a valid approach to separate the problem into two problems where you treat each as a constant, and then sum the answers.

[@phiefer3]

@@xinpingdonohoe3978This actually somewhat makes sense if you consider it as a total derivative of a 2 variable equation. When working with multivariable equations you can take a partial derivative with respect to one of the variables by treating the other variables as if they were constants. And then the total derivative is essentially the sum of all the partial derivatives. So if we change x^x into something like x^t and then take the partial derivatives, we'd get dy/dx = tx^(t-1), and dy/dt = ln(t) x^t, so the total derivative is the sum of these: ln(t)x^t + tx^(t-1), and then we change the t's back into x's (and do the bit of simplification) and we get the derivative for x^x. It's also worth noting that this works for other simpler derivatives as well, for example y=x^2, we can instead write this as x*x, and change one of them so we have x*t. Now dy/dx = t, and dy/dt = x, add them together and we get x+t, change the t back to x and we have x+x = 2x, which is indeed the derivative of x^2. Or y=2x = x+x turn that into x+t, dy/dx = 1, and dy/dt = 1, add them together and we get 2, which is the derivative of 2x. Generally, for single variable functions this isn't particularly useful, as it usually just makes more work to differentiate a function in parts like this. x^x seems to be an odd exception where this would be useful.

[@IscAst4]

Dios te bendiga

[@tomctutor]

Ha, ha! Now find the derivative of the most general exponential form d/dx { (f(x))^ g(x) } ? 😅

[@damiangowor1404]

At last, it's legit.

[@sumcarson]

Cool!

[@reckless_r]

🤦

[@Modie]

When I clicked on the video, I expected a trick question. Because x^x is not actually derivative. You can only take the derivative for x>0.

[@lotaniq4449]

x is a constant so d(x)/dx is 0

[@rv706]

Because it's neither a power function nor an exponential funcion: case closed... 🤷🏼

[@Johannes_Seerup]

You can

[@user-zu4mr9bf6j]

Thank😀😀😀