Derivative of ln x

In this video, I show that the derivative of ln(x) is 1/x using the definition of the derivative. In my opinion, this way is more elegant than the classical way, which is to use that ln(x) is the inverse of e^x. Enjoy!

[Aviationlover-belugaxl]

I made a subreddit for your channel, r/DrPeyam.

[nchoosekmath]

Nice video!

I had a conversation with a friend before but it really depends on the definition of ln. In one textbook, ln was defined to be the indefinite integral of 1/x from 1 to infinity. From there, e^x is defined to be the inverse of ln. Using this definition, derivative of ln is 1/x by definition and fundamental theorem of calculus.

[rigorless]

i think this is how early transcendental textbooks do it.


- if youve already defined e by the limit definition before calculus (early transcendentals), then you would use this to get the derivative of lnx, and then use 'derivative of inverse' inverse trick to get the derivative of e^x. this is how i learned it in school.
- if you havent defined e before calculus (late transcendentals), then you can define lnx as integral of 1/x, and then use the 'derivative of inverse' to get e^x. this is actually how it was done historically.

[AdityaKumar-ijok]

why are you differentiating ellen?
how does ellen behave when you differentiate her?

[MrCigarro]

This prove is, indeed, more elegant. Thanks.

[StreuB]

That was absolutely beautiful. We were never shown this in Calc 1, we were just told it was true and that was that. Thank you Dr. Peyam!!

[madhuragrawal]

Isn't a better proper definition of exp (and by extension, ln) the functional equation and not the limit or integral?

[Mrwiseguy]

You can also do ln(x)=f(x), x=e^u => ln(e^u)=f(e^u) => u=f(e^u). Take derivative and apply chain rule. du/dx = f'(e^u). Recalling that x=e^u, take the derivative of both sides and you get 1 = e^u * du/dx => 1 = x * du/dx. Solve for du/dx and you get du/dx = 1/x. Plug the result back and you get 1/x = f'(e^u) which is 1/x = f'(x).

[Czeckie]

yep, you don't need any differentiation to get the limit. The limit is quite often the definition for e. The validity of such definition is established by easy ratio tests and squeezes

[googleuser]

U r like a Mariner's compass for students, which guides them on non-conventional paths but always lead to the correct result and always with a beautiful smile :-) on his face.
Again an elegant video with a different approach of find'g the result.
BTW, is this comment section the right place for the struggling YT learners to ask you some doubts/queries?

'M struggling with this problem:
If (x->0)lim[1+x+f(x)/x]^(1/x) = e^3, then find the value of ln[(x->0)lim{1+f(x)/x}^(1/x)] is

[ChefSalad]

But what about if we allow the domain to be ℂ. Or even x<0. I′m not satisfied with this x>0 stuff

[ravitsharma]

what about ln of πn?? XD


It's always fun to use the definition of the derivative to discover hidden hidden properties of functions. Flammable Maths applied it to a^x to demonstrate that ln(a) is equivalent to lim (a^h - 1)/h as h approaches 0... Cool video!! :)))

[Quadraticmula]

Please address the arbitrary choice of raising to the (1/X). Why not put (x^2/h)*(1/x^2)? Why or how did you know to choose 1/x? Please & thank you.

[MrRyanroberson]

i prefer to go even more directly from ln((1+h/x)^(1/h) to y=x/h, ln((1+1/y)^(y/x)) to ln(e^(1/x)) to 1/x, though that also uses ln(e^x)=x.

[dolevgo]

7:55 you could use the fact that x=e^lnx
by differentiating both sides you get 1=(lnx)'*e^lnx = (lnx)'*x
1=(lnx)'*x => (lnx)'=1/x

[dominikstepien]

Isn't this limit the definition of the number e?

[phyarth]

Nice and in the end pushed to far, h→0 limit nicely converges inside natural logarithm into constant e=2, 718.., but to rise ln by 1/x, to prove that derivative of ln(x)=1/x is "creative" math.

[newtonnewtonnewton]

Thanks a lot D peyam السلام عليكم /عيد اضحى مبارك

[johannesh]

But does it really converge for h (overbar) - > - infinity

[VibingMath]

Learnt a new way of d/dx lnx
Thanks Dr Peyam!