d/dx (x^x) *BAD MATH*

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[BriTheMathGuy]

Check out the Full Vid for more Bad Math! (That works)

[sadChris]

“add the two to get partial credit“ 😂
i love that 😂

[stardust-rz]

It's valid, more precisely
df(g(x), x)/dx = ∂f/∂x + ∂f/∂g dg/dx

[venkataseshasai]

x^x=y
xln(x)=ln(y)
Differentiate on both sides
(x/x)+1.x=(1/y)dy/dx
y( l+lnx)=dy/dx
[y=x^x]
(x^x)(l+lnx)=d(x^x)/dx

[gautamsingh]

This actually isn't bad maths and its a real method where you first keep the base constant and the power variable and differentiate and then keep the base variable and the power constant and differentiate...then add the both of them

[fullfungo]

This is not bad math.
This is just a more intuitive way of doing multi variable calculus, and taking total derivative.

[marshmello]

we can write x^x as e^xlnx, Dx^x =De^xlnx = e^xlnx. D(xlnx) =x^x(1+lnx)

[kgyo]

This method leads to

When the real derivative is

Therefore if f'(x)=1 and g'(x)=1, this is true
So it can be used for any (x+k)^(x+m)

[MC_Transport]

Correct method: Logarithmic differentiation
Base is a variable -> Can't use the exponential rule
Power is a variable -> Can't use the power rule

[Mebasically]

That's not wrong math, it's a correct way to do it-
d/dt [f(t, t)] = ∂f/∂x dx/dt+∂f/∂y dy/dt =
(Since x=y=t, dx/dt=dy/dt=1)
= ∂f/∂x+∂f/∂y
This is exactly what he did...and it works in general (which is kind of nice).

[byashverma]

It's a very common trick taught by Indian teachers to students in case if they forgot og method

First take the bottom x as variable and power x as constant
Next take the power x as constant and power x as variable

Add them

[Johnny-twpr]

Its actually adding up the x and y compenent of the gradient(nabla)f(x, y) and set y=x

[mumujibirb]

One time I had do x^x^x and you just re-express as powers of e and chain and stuff to the end.

[Nightlife_Offical]

Does x^x not just differentiate to x x^x-1 = x^x again? Or does the normal rule not apply in this instance and why doesn’t it apply if so?

[lackdejuranez]

Isn't this just like using partial derivatives?

[kgyo]

This can used for any m and k when calculating the derivative of (x+m)^(x+k)

[dienosorpo]

Wasn't this the original derivation formula?

I was taught to do it this way

[keshav]

Bro this is using partial derivatives not tricks

[Al_Pal]

I only know the way where u change it to e^ln(x)^x=e^xln(x) and then u diff from there