How to Differentiate x^x ? [2 Different Methods]

There are 2 different ways to take the derivative of x^x, which are implicit differentiation, and the chain rule. In this video, we will be solving for the derivative of y=x^x by using these two methods. For the implicit differentiation, we first take the natural log on both sides of the equation, and we are able to apply implicit differentiation to solve for the derivative. For the chain rule, we also take the natural log on both sides, but the difference is that instead of taking the derivative on both sides directly, we first rewrite the equation from logarithmic form into exponential form, so that we can apply the chain rule.
Implicit Differentiation Explanation:
→ • Learn Implicit Differe...
Chain Rule Explanation:
→ • Understand Chain Rule ...
TIMECODES:
0:00 Intro
0:14 First Method - Implicit Differentiation
2:17 Second Method - Chain Rule
4:12 Outro

[@itsale5918]

You can also rewrite x^x as e^(xlnx) since this is in exponential form. From there, the term e^(xln(x)) stays the same and you multiply it by the derivative of xln(x) (use product rule). On the right side, you get e^(xlnx) * (1+lnx) and on the left side, the derivative of y with respect to x is 1 dy/dx. Rewrite e^(xlnx) as x^x and you get the same answer as here.

[@YeahMathIsBoring]

Exactly! Check out the second method that I've done in 2:22

[@user-xb1ql9xz2t]

I didn't know we could solve it using the chain rule . Thanks , pal !

[@kunlong-vp2qx]

good job! thank you!

[@YeahMathIsBoring]

Thanks!

[@anotherelvis]

Or with partial derivatives:
d/dx x^x = d/dy y^x |y=x + d/dy x^y |y=x
= ln(y)*y^x |y=x + y*x^(y-1) |y=x
= ln(x)*x^x + x*x^(x-1)
= (ln(x) + 1) *x^x

[@adw1z]

Awesome, these are two fundamental methods everyone should know about
Now differentiate x^^3 = x^(x^x) B)

[@user-nd7th3hy4l]

dy/dx=(lnx+1)x^x

[@mathsfamily6766]

love it

[@tomvitale3555]

Pure beauty!

[@YeahMathIsBoring]

Thanks!

[@josepherhardt164]

I know it doesn't work, but I'd've liked to have had at least a brief discussion about why using the exponent function derivative doesn't work, e.g.,
we know y = x ^ n has derivative y' = nx^(n-1)
If we attempt this with y = x^x, we get y' = x(x^(x-1)) * dx/dx = x^(x-1+1) * 1 = x^x
I guess my question here is, _exactly what rule is it we're violating when we attempt this?_

[@zachberman3423]

when you differentiate x^n, n must be a constant in order for the power rule formula to apply. In this case, x is raised to a variable power (x) and therefore this rule does not apply.

[@user-nk7iq9ej9m]

Easy, it's x.x^(x-1), done.

[@Spyro101]

Why is natural log used Instead of normal log?

[@josepherhardt164]

Because to take the derivative of y = k ^ x, you'd need to convert the base k to e ^ ln(k) anyway before you could take the derivative. When you choose e = k, the term ln(k) = 1 and drops out. I know this doesn't sound like an actual explanation--though it is--it's one of those mathematical things that you need to contemplate for a while, like a navel, to understand.

[@themisfowl1333]

now integral?

[@alvarotriguerosalonso7630]

you don't

[@garfungled7093]

Sphi(x)

[@ianweckhorst3200]

I would gladly show you how to derive x^x^x^x, but integrating it is a different story, I could even show you the pattern behind deriving larger and larger power towers, never integrate

[@ianweckhorst3200]

From an observation, if the integral is p(x), p(x)/x is weirdly close to (x/1.5)^x/1.5

[@ianweckhorst3200]

Not exactly though, and it seems to spit out a bunch of irrationals, can’t prove they’re irrational though

[@victor1978100]

There is a third method. Differentiation of function of two variables.

[@ianthehunter3532]

could you explain further what you mean?

[@victor1978100]

@@ianthehunter3532 We can treat each of the x's as two different variables. Then treat those two variables as two different functions. The derivative of a function of two variables is equal to the sum of two partial derivatives. Sorry for my English.

[@ianthehunter3532]

​@@victor1978100 Right, so you make only the first x a constant, then only the second x, and you equal the derivatives?

[@victor1978100]

@@ianthehunter3532 x*x^(x-1)(here we multiply by x and divide by x, so we get x^x)*x '(which equals 1)+x^x*ln(x)*x ' (which also equals 1) and we get x^x*(1+ln(x)) This method works for any two functions. For example, cos(x)^ln(x).

[@ianthehunter3532]

​@@victor1978100 Apologies, but still I don't quite get the way you are solving it. Maybe this attempt to solve with what you told help can help explain further? i . imgur . com / LZSFrD9.png I added spaces to the link.

[@emmanuelisaac5988]

nice

[@YeahMathIsBoring]

Thanks!

[@dacooooord2114]

I just automatically assumed x=exp(ln(x)) and we just use the rules of derivatives to get y'= ln(x)exp(ln(x)×x)=ln(x)x^x but you said that it was (1+ln(x))x^x i just cant figure where im wrong

[@zachberman3423]

In order to use this formula for the derivative of a^x, a must be a constant. In this case, the base is another variable x and therefore this formula does not apply

[@gelbkehlchen]

1.Solution:
y = x^x with 0

[@telmanm6987]

the solution is x=1,y=1

[@TramNguyen-pk2ht]

Yes but no

[@rennoc6478]

That’s one of the solutions