i simply love this channel. You work so hard to make us understand these things, it may look like a small and simple topic but its actually not and we understand so much more because of the effort you put into these vids. Its arbitrary how you only have 21 (22 now :) subscribers. I really hope creators like you grow
@YeahMathIsBoring
10 ай бұрын
Thank you so much for your support!
@Fire_Axus
3 ай бұрын
your feelings are irrational
@anglelee632
10 ай бұрын
Not gonna lie.... i love it you make it really easy for us to understand
@YeahMathIsBoring
10 ай бұрын
Thanks!
@Fire_Axus
3 ай бұрын
your feelings are irrational
@roulio7908
8 ай бұрын
Does this work ? y = ln(x) x = e^y dx/dy = e^y = x dy/dx = 1/x
@YeahMathIsBoring
8 ай бұрын
It seems like you tried to differentiate it with respect to y, and taking the derivative of e^y we will still be getting e^y, which is basically equivalent to x. Then, we flip the dx/dy to dy/dx, so we end up with 1/x. Based on your mathematical workings, it should be no problem. Anyway, that was a nice approach to this question. Well done!
@marianoses4282
6 ай бұрын
You are making circular reasoning. Aren't you? When you calculate the derivative of ln(x) you use the derivative of the exponencial función and viceversa. I mean, in another video,when you calculate the derivative of the exponencial función you use the derivative of the logarithmic función.
@filipeoliveira7001
5 ай бұрын
Well not really cuz you can get the exponential derivative from the standard definition of a derivative
@sunhouse8616
5 ай бұрын
i outlined a noncircular derivation in my own comment.
@fyodorvasilev1964
4 ай бұрын
It's not circular. We have a definition(s) of exp, from which its derivative is derived. ln is defined through exp. So it's an implication, not a circle. You can also use the formula for the derivative of an inverse. Once again, it's simply an implication of a definition.
@c01d_h4nds
8 ай бұрын
Who tf says lawn X? 🥴
@bizw
8 ай бұрын
AI does
@Eggcellent_Sandwich
8 ай бұрын
I was taught this at my high school, but I'm in Australia so idk if it's a country thing
@prabhakarsingh6821
5 ай бұрын
A lot of people
@c01d_h4nds
5 ай бұрын
@@prabhakarsingh6821 learn something new everyday
@le24_qr6nod
4 ай бұрын
I do
@Lee-bs8oq
11 ай бұрын
You made it clear and easy to understand
@nuggetlover9431
5 ай бұрын
We can use the limit definition to...
@xylh5085
10 ай бұрын
I was low key hoping this would be a spicy and unhinged usage of the limit definition of the derivative. Regardless, that's an elegant proof
@YeahMathIsBoring
10 ай бұрын
This is just an another way of thinking for taking the derivative of ln x by rewriting it into exponential form. Anyway, I'd try to make a video soon for the first principle of differentiation
@adw1z
9 ай бұрын
[ln(x)]’ = lim h->0 [ln(x+h) - ln(x)] / h = lim h->0 ln(1+h/x) / h (use Taylor Series for ln) = lim h->0 [h/x - h^2/2x^2 + O(h^3/x^3)] / h = lim h->0 1/x + O(h/x^2) = 1/x [Note: we derive the Taylor series for ln without using the assumption that we know it’s the derivative as follows: if y = ln(1+x), x = e^y - 1 = y + y^2 / 2 + y^3 / 6 + .. (*) Suppose ln(1+x) = a0 + a1 x + a2 x^2 + … for ai constants to be found. ==> x = (a0 + a1 x + …) + (a0 + a1 x + …)^2 / 2 + … We equate coefficients of powers of x to find that: a0 = 0 , ai = (-1)^(i-1) / i for i >= 1 Thus, ln(1+x) = x - x^2 / 2 + x^3 / 3 - … simply by inverting the infinite series (*) ] I probably overcomplicated it lol, there probably is a much easier way to calculate the limit but I’m too tired to figure it out
@carultch
9 ай бұрын
@@adw1zUsing Taylor series would be circular reasoning, since the Taylor Series terms are ultimately derived from the derivative.
@adw1z
9 ай бұрын
@@carultch I found the Taylor expansion using series inversion, not finding any derivatives
@aadityakiran07
5 ай бұрын
ur so underated omg
@gcb642
3 ай бұрын
next video -> why d(e^x) = e^x*dx
@AllanPoeLover
2 ай бұрын
0:39 How come d/dx(e^y) become e^y(dy/dx) ? Where did the y come from?
@williamlong63
2 ай бұрын
That's the chain rule.
@minhhungle7488
6 ай бұрын
If u try to differentiate a^x by definition, you will encounter a limit expression which is not related to x. You differentiate it with respect to a and you will get 1/a
@sailorrev1
6 ай бұрын
How does this work using the derivative definition?
@NotSoChillBozo
6 ай бұрын
Basically you have Lt(h->0) [ln(x+h)-lnx]h Use the property of log to get ln[(x+h)/x]/h (limit is still in place) Now (x+h)/x simply equals 1+h/x So multiply divide by 1/x to get (1/x)*ln[1+h/x]/(h/x) Use limit a->0 ln(1+a)/a=1 Here, a=h/x This you're left with 1/x
@abhirupkundu2778
7 ай бұрын
that last step was unnecessary. As soon as we got 1/e^y, e^y was = x, so u could've just put that
@ramzyalexandre5725
10 ай бұрын
Wonderful
@YeahMathIsBoring
10 ай бұрын
@alejandropulidorodriguez9723
7 ай бұрын
splendid
@japethspeaketh7034
5 ай бұрын
You say Laun, I say Alan.
@gordontan2092
10 ай бұрын
why d/dx e^y becomes e^y dy/dx
@YeahMathIsBoring
10 ай бұрын
It is actually an Implicit Differentiation of e^y with respect to x. First we differentiate e^y in terms of y, we will still be getting e^y. Then, multiply by dy/dx because we are taking the derivative of the terms that has y in it, so we end up with e^y (dy/dx). Feel free to check out the video if you wanted to learn more about Implicit Differentiation: kzitem.info/news/bejne/jnqhro2OiH2olaA
@YeahMathIsBoring
10 ай бұрын
In other words, when we are differentiating something that has y in it, we differentiate it normally just like we always do, but we multiply it by dy/dx.
@gordontan2092
10 ай бұрын
@@YeahMathIsBoring Ohh i understand thanks a lot
@YeahMathIsBoring
10 ай бұрын
@@gordontan2092 Glad that you've understood. You're welcome!
@makramaarid6598
3 ай бұрын
1:07 Why all that talk 🤣 we know that eⁿ = x remplace it
@Harwey-lz4gp
6 ай бұрын
If you use this method with: y = log,a,(x) You also get 1/x but thats wrong
@YeahMathIsBoring
6 ай бұрын
In fact, we wouldn't get 1/x if we use this method for differentiating y=log_a(x). You see the first step we take is to rewrite the equation into exponential form, then finding the derivative of this equation using implicit differentiation. In the case of y = log_a (x), if we rewrite it into exponential form as well, we actually end up with a^y = x, which requires a slightly different solution, in which we will be getting 1/(x * ln a). For more information, you made check out the following video: kzitem.info/news/bejne/15t_zWSmppteaoYsi=fDt96aIjUNWZrFVB
dy/dx is the derivative of y with respect to x. d/dx is a derivative operator (with respect to x), that tells you to differentiate. Like how + is the addition operator
@afonsopina3068
6 ай бұрын
@@wyldcat9396 thanks❤️
@mathsfamily6766
5 ай бұрын
very nice video! thanks for sharing
@lawrencelawsen6824
3 ай бұрын
I love this!
@sameermansour1659
2 ай бұрын
so simple and beautifull explanation !
@abinadvd
2 ай бұрын
Not really, it is when you know "everything" you have this. It depends how you build the function log(x) At start , you have exp(x) which is a function who exp(x)'=exp(x)... You are happy, it is the only one like so. You see exp(x)=e^x, it gives you e=2.718... A) One way for log definition Now , you are searching primitive for x^n where n is an integer (positive or negative). It is for polynomials primitives and for 1/x^n if n>0 . (x^n)'=n*x^(n-1) , You have primitve too any n : if f(x)=x^n ... F(x)=1/(n+1)*x^(n+1). Your are very happy , for any n you have a primitive. But you have a problem , it is not ok for n=-1. You dont have a primitive for 1/x with this method ! So you say : OK, I will call log(x) the primitive of 1/x (the good one with log(1)=0). In fact HERE log(x)'=1/x because it is définition of log(x) !!! Then proove exp(x) and log(x) are reciprocal functions... B) Second way : the way used in this video. You say log(x) is the reciprocal of exp(x) : definition of log(x) Then you calculate log(x)'=1/x. Conclusion : there is 2 ways possible to create log(x) : reciprocal of exp(x) or primitive of 1/x. It is a choice. The result is the same.
@jiamei_426
11 ай бұрын
why moving ln to the other side becomes e to the power of something?
@YeahMathIsBoring
11 ай бұрын
'ln' is basically the log of base e. When we move it to the other side, which means we switch it from logarithmic form into exponential form. For example, think of that we have log base 2 of 8, which the answer is equal to 3. It is exactly the same as 2^3 = 8.
@jiamei_426
11 ай бұрын
Owhhh, now i get it. Thanks anyway!
@YeahMathIsBoring
10 ай бұрын
@@jiamei_426 Anytime!
@williamlong63
2 ай бұрын
because they are not really moving ln to the other side. It was a figure of speech.
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