THE CONFUSING DERIVATIVES

So did the derivatives of tanh^-1(x) and coth^-1(x) and found out they are the same! But.... are they???

[nathanisbored]

so basically, its like saying ln(-x) and ln(x) have the same derivative, but they correspond to different domains of 1/x

[AndDiracisHisProphet]

8:00
"one plus one is
the more you know

[Jono]

If you work with the complex logarithms then you actually can argue that those two functions differ by a constant complex number because 1/2 ln( (1+x)/(1-x) ) = 1/2 ln( (x+1)/(x-1) ) + 1/2 ln(-1).

[not_vinkami]

Actually... tanh^-1(x) and coth^-1(x) can still be said as 'off by a constant', it is just that the constant is πi/2

[ssdd]

actually, there is a constant of 1/2((2n+1)πi) where n is an integer

edit: oops i forgot about the 1/2

[DiegoTuzzolo]

They are indeed off by a constant, which is pi*i/2

On the first expression, factor the -1 from the denominator:
1-x = (-1)*(x-1)
so it becomes:
1/2*ln( (-1)*( x+1/x-1))
using the ln properties, this becomes:
1/2*(ln(-1) + ln(...))
ln(-1) = pi*i
Thus the expression equals:
1/2(pi*i +ln(...)= cosh^-1(x) +pi*i/2

[scarletevans]

Nice video!
Always good to hear about things that work similarly, but on different domains! 🙂
Especially that many people forget to consider differences in domains.
It would be great to see more content like this one! 🙂

[XanderGouws]

What about when we allow things to be complex?

If we let 1/2 ln((1+x)/(1-x)) = 1/2ln((1+x)/(x-1)) + C, we get C = ln(i) = i π/2.
So technically, we can say that they are the same function + a constant, just that the constant is complex.
Right?

Edit: If we're looking for a set of all constants that work I think it's C = (k + 1/2) π i, where k is any integer.

[calyodelphi]

I love that you've been covering the hyperbolic trig functions so much recently. It's been helping to demystify them. :)

[Debg]

Very interesting indeed. I have never faced this problem because I have always worked with these functions in the complex world. But in the real world the indefinite integral of 1/(1-x^2) would be a piecewise function right?

[krischan]

The mystery will unfold if the graphs of the inverse functions of tanh and coth are are drawn in the same coordinate system.

[sthubbar]

Amazing! You blew my mind. It is great to see these simple examples that shatter my confidence that I understand math. Thank you.

[KingGisInDaHouse]

At 1:15 you can just take d/dy to get dx/dy, then you just flip it over to get dy/dx. it simplifies real nicely with the hyp trig identities.

[BigDBrian]

different domain. ln can't be negative. The numerators in the ln are identical but the denominators are negatives of each other because -(1-x)= -1+x = x-1.
This means that whenever one is positive, the other is negative and thus not real.

[mattmackay]

THAT ... WAS .... AWESOME.. your channel is my new favorite! Keep up the great work

[benjaminbrady]

Wait though, if you include complex numbers then are the graphs the exact same?

[housamkak]

no, I'm studying for my linear algebra final and I ended up here. but great video though.blessed efforts.

[helloitsme]

The constant is there but its a complex constant right?

[quocanhnguyenle]

Hi, bprp!
So I was playing around with angles, and I found out that there exists an angle α between 0 and 90 degrees such that cos(α) = tan(α).
And, the neat part is, sin(α) is exactly 1/φ (or you can say, csc(α) = φ).
(φ is the golden ratio)
Pretty cool, isn't it? Maybe you should make a video about this!

[Sky]

So I guess you can basically say.. (Its propably wrong the way I say it in english).. the tanh^-1(x) being the steady continuation of coth^-1(x) to define a function all over the real numbers?