To proof the derivative of lnx = 1/x

[drillby]

Hello all, I have a question about proofing the derivative of lnx = 1/x.
Quoting from Kumon,

I understand from top to mid until I read the one that I circled with a blue pen.
I don't understand how we can get to the conclusion that the lim k-> 0 ln(1+k)^1/k = 1 just from the lim t->~ (1+1/t)^t = e.
Sorry for the bad quality picture and sorry if I posted in the wrong section.
Thank you.

 

[pka]

Hello all, I have a question about proofing the derivative of lnx = 1/x.
Quoting from a trusted source,
I understand from top to mid until I read the one that I circled with a blue pen.
I don't understand how we can get to the conclusion that the lim k-> 0 ln(1+k)^1/k = 1 just from the lim t->~ (1+1/t)^t = e.
Sorry for the bad quality picture and sorry if I posted in the wrong section.

First, we have no way to know about your source. All of this depends upon how that source defines the logarithm function. We don't know that. You have to tell us.

 

[drillby]

First, we have no way to know about your source. All of this depends upon how that source defines the logarithm function. We don't know that. You have to tell us.

Sorry, I never thought that it would be that complicated. But I also don't get what you mean All I can say is that it is my Kumon sheet about advanced differentiation. I can take a photo of the sheet but I was concerned about the copyright so I rewrite exactly the same on another piece of paper. Unfortunately, my teacher can't teach me.
I watched Khan Academy's video too
https://www.khanacademy.org/math/di...atives-of-ex-and-lnx-dc/v/proof-d-dx-ln-x-1-x
but I don't understand it too

 

[Deleted member 4993]

\(\displaystyle \displaystyle{lim_{x\to 0}{ ln(1+x)^{\frac{1}{x}}}}\)

\(\displaystyle \displaystyle{= \ ln(lim_{x\to 0}{(1+x)^{\frac{1}{x}})}}\)

= ln(e) = 1

 

[pka]

\(\displaystyle \displaystyle{lim_{x\to 0}{ ln(1+x)^{\frac{1}{x}}}}\)

\(\displaystyle \displaystyle{= \ ln(lim_{x\to 0}{(1+x)^{\frac{1}{x}})}}\)

= ln(e) = 1

How do we know that \(\displaystyle \displaystyle{lim_{x\to 0}{ ln(1+x)^{\frac{1}{x}}}}=e~?\)
How do we know that \(\displaystyle \log(e)=1~?\)

That is my whole point. The augment is circular.

 

[drillby]

\(\displaystyle \displaystyle{lim_{x\to 0}{ ln(1+x)^{\frac{1}{x}}}}\)

\(\displaystyle \displaystyle{= \ ln(lim_{x\to 0}{(1+x)^{\frac{1}{x}})}}\)

= ln(e) = 1

Whoa is it possible to switch place between the lim and ln?
before it was lim(ln(blablabla)) now it becomes ln(lim(blablabla))
is there a rule for switching that ??
and moreover the lim x->0 is different with lim t-> ~ since the value is different, one is 0 and the other is ~?? why we can replace the
lim x-> 0 with e then
sorry for the problem >.<

oh I think I know because

is it?? but I still don't get why we can switch the lim and ln

 

[ksdhart2]

Well, I'll start with the easier of your questions - yes you're correct that it's to do with the fact that one variable goes to infinity and the other goes to 0. If you wish, you can think of it as a variable substitution where u=1/t. As t approaches infinity, u approaches 0, so:

\(\displaystyle \displaystyle \lim _{t\to \infty }\left(1+\frac{1}{t}\right)^t\) has the same value as \(\displaystyle \displaystyle \lim _{u\to 0}\left(1+u\right)^{\frac{1}{u}}\)

The answer to your other question is a bit more complicated, but the answer there is also yes. It doesn't matter if you put the natural logarithm outside or inside the limit, you'll get the same value. You can try it yourself on a simpler limit and see it works:

\(\displaystyle \displaystyle \lim _{x\to 3}\left(ln\left(x^2\right)\right)=ln\left(\lim _{x\to 3}\left(x^2\right)\right)=ln\left(9\right)\)

This trick of "pulling out the function" works for many other functions too. Taking the limit as a whole to any power is the same as taking argument of the limit to that power, and trig functions (sine, cosine, etc) work too.

 

[pka]

Whoa is it possible to switch place between the lim and ln?
before it was lim(ln(blablabla)) now it becomes ln(lim(blablabla))
is there a rule for switching that ??
is it?? but I still don't get why we can switch the lim and ln

You have perfectly shown why I questioned the basis of this question.
If \(\displaystyle f\) is continuous at \(\displaystyle b\) and \(\displaystyle \displaystyle{{\lim _{x \to a}}g(x) = b}\) then \(\displaystyle \displaystyle{{\lim _{x \to a}}f\left( {g(x)} \right) = f\left({{{\displaystyle{\lim }_{x \to a}}g(x)}} \right)}\).
You may not know that theorem. We have no way to know that. Do you see now why this is complicated?

 

[HallsofIvy]

lim f(g(x))= f(lim g(x)) if and only if f is continuous. That is, basically, the definition of "continuous".

 

[drillby]

Well, I'll start with the easier of your questions - yes you're correct that it's to do with the fact that one variable goes to infinity and the other goes to 0. If you wish, you can think of it as a variable substitution where u=1/t. As t approaches infinity, u approaches 0, so:

\(\displaystyle \displaystyle \lim _{t\to \infty }\left(1+\frac{1}{t}\right)^t\) has the same value as \(\displaystyle \displaystyle \lim _{u\to 0}\left(1+u\right)^{\frac{1}{u}}\)

The answer to your other question is a bit more complicated, but the answer there is also yes. It doesn't matter if you put the natural logarithm outside or inside the limit, you'll get the same value. You can try it yourself on a simpler limit and see it works:

\(\displaystyle \displaystyle \lim _{x\to 3}\left(ln\left(x^2\right)\right)=ln\left(\lim _{x\to 3}\left(x^2\right)\right)=ln\left(9\right)\)

This trick of "pulling out the function" works for many other functions too. Taking the limit as a whole to any power is the same as taking argument of the limit to that power, and trig functions (sine, cosine, etc) work too.

I got it thanks !

You have perfectly shown why I questioned the basis of this question.
If \(\displaystyle f\) is continuous at \(\displaystyle b\) and \(\displaystyle \displaystyle{{\lim _{x \to a}}g(x) = b}\) then \(\displaystyle \displaystyle{{\lim _{x \to a}}f\left( {g(x)} \right) = f\left({{{\displaystyle{\lim }_{x \to a}}g(x)}} \right)}\).
You may not know that theorem. We have no way to know that. Do you see now why this is complicated?

lim f(g(x))= f(lim g(x)) if and only if f is continuous. That is, basically, the definition of "continuous".

Sorry but what does f means?? is any function like ln, sin, cos, etc??

 

[stapel]

Sorry but what does f means?? is any function like ln, sin, cos, etc??

Yes, "f" is any function. Just as specific functions receive specific names "ln()", "sin()", etc, so also generic functions can receive generic names, like "f()", "g()", etc.

 

[drillby]

Yes, "f" is any function. Just as specific functions receive specific names "ln()", "sin()", etc, so also generic functions can receive generic names, like "f()", "g()", etc.

Ok thanks

 

[HallsofIvy]

Getting back to the original question, determining the derivative of ln(x), how you find the derivative of any function depends strongly on how you define that function! So exactly how are you defining "ln(x)"?
One definition that I particularly like, and is showing up more and more in Calculus books is \(\displaystyle ln(x)= \int_1^x \frac{1}{t} dt\). (We already have an integral for all \(\displaystyle x^n\) except n= -1. This fills that gap.) With that definition the "Fundamental theorem of Calculus" gives immediately that \(\displaystyle \frac{d(ln(x)}{dx}= \frac{1}{x}\).

Perhaps more common is to define \(\displaystyle ln(x)\) as the inverse function to \(\displaystyle e^x\). That is, if \(\displaystyle y= ln(x)\) then \(\displaystyle x= e^y\). The derivative of that is \(\displaystyle \frac{dx}{dy}= e^y\) so the \(\displaystyle \frac{dy}{dx}= \frac{1}{e^y}= \frac{1}{x}\).

 

[pka]

One definition that I particularly like, and is showing up more and more in Calculus books is \(\displaystyle ln(x)= \int_1^x \frac{1}{t} dt\). (We already have an integral for all \(\displaystyle x^n\) except n= -1. This fills that gap.) With that definition the "Fundamental theorem of Calculus" gives immediately that \(\displaystyle \frac{d(ln(x)}{dx}= \frac{1}{x}\).

I absolutely agree that is the best approach. I am convinced the it first appeared in 1973 calculus text be Gillman & McDowell (I still that edition is the best calculus text ever). I had the great fortune to have been is a small group discussion with Gillman about his reasoning to push for a single definition for \(\displaystyle \log(x)\). But in that text the definition requires that \(\displaystyle x>0\).

 

[HallsofIvy]

I absolutely agree that is the best approach. I am convinced the it first appeared in 1973 calculus text be Gillman & McDowell (I still that edition is the best calculus text ever). I had the great fortune to have been is a small group discussion with Gillman about his reasoning to push for a single definition for \(\displaystyle \log(x)\). But in that text the definition requires that \(\displaystyle x>0\).

Yes, of course. Since 1/x is unbounded in a neighborhood of 1/x, that integral can not be extended to or past x= 0.