C0balt
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Hello
We were doing FP1 topic on calculus and the teacher was about to do an example using 1/x but he was like oops it's an exception and the integral of 1/x is actually ln x.

I have no idea why this is true (my teacher didn't know/couldn't explain) and I have searched on Google but explanations there are incomprehensible for me.
Is there any way for me to understand this thing having only done C1 and half of FP1 so far? Will I be able to understand it by the end of A2 FM?


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TenOfThem
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(Original post by C0balt)
Hello
We were doing FP1 topic on calculus and the teacher was about to do an example using 1/x but he was like oops it's an exception and the integral of 1/x is actually ln x.
You will learn about differentiating natural logs and exponentials in C3/4
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Prince Philip
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(Original post by C0balt)
Hello
We were doing FP1 topic on calculus and the teacher was about to do an example using 1/x but he was like oops it's an exception and the integral of 1/x is actually ln x.

I have no idea why this is true (my teacher didn't know/couldn't explain) and I have searched on Google but explanations there are incomprehensible for me.
Is there any way for me to understand this thing having only done C1 and half of FP1 so far? Will I be able to understand it by the end of A2 FM?


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I'm not sure what kind of explanation you're looking for

E.g. are you happy that the integral of x^2 is x^3/3? What explanation has made you accept this?
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DFranklin
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(Original post by TenOfThem)
You will learn about differentiating natural logs and exponentials in C3/4
But will you get told anything more than "the derivative of log x is 1/x but it's too difficult to explain why"? [Genuine question - I don't know how it's taught these days].
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tombayes
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(Original post by C0balt)
Hello
We were doing FP1 topic on calculus and the teacher was about to do an example using 1/x but he was like oops it's an exception and the integral of 1/x is actually ln x.

I have no idea why this is true (my teacher didn't know/couldn't explain) and I have searched on Google but explanations there are incomprehensible for me.
Is there any way for me to understand this thing having only done C1 and half of FP1 so far? Will I be able to understand it by the end of A2 FM?


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Ok here is a 'basic i.e. non-formal' idea: the derivative of e^x is itself. i.e. let y=e^x then \frac{dy}{dx}=e^x=y

If we rearrange  \frac{dy}{dx}=y we see that \frac{1}{y}dy=dx now integrating both sides gives \int \frac{1}{y}dy=x+c where c is constant but recall e^x=y \implies x=ln y

so \int \frac{1}{y}dy=ln(y)+c
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tombayes
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(Original post by notnek)
E.g. are you happy that the integral of x^2 is x^3/3? What explanation has made you accept this?
if you view integration as anti-differentiation then this is easy to 'accept' since they would have covered first-principles differentiation. You really cannot ask for more as this level.
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Prince Philip
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(Original post by DFranklin)
But will you get told anything more than "the derivative of log x is 1/x but it's too difficult to explain why"? [Genuine question - I don't know how it's taught these days].
The most common way I've seen teachers/textbooks introduce it at C3 is by assuming the derivative of e^x and using y=ln(x) --> x=e^y with dx/dy = 1/(dy/dx).
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poorform
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At A-Level I was just told that it is what it is with no real reasoning.

I still don't really understand why it is but I'm starting a calc modulus after christmas and I know for a fact it's on there.
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DFranklin
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(Original post by notnek)
The most common way I've seen teachers/textbooks introduce it at C3 is by assuming the derivative of e^x and using y=ln(x) --> x=e^y with dx/dy = 1/(dy/dx).
And I assume you prove the derivative of e^x by using the derivative of ln x and using similar reasoning...

[Obviously, the involvement of 'e' is a massive spanner in the works at A-level, since it's a completely unfamiliar number and the definitions of it are pretty intractable (\sum 1/n! is fairly easy to understand, but how the heck you work with the result is less so...)]
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atsruser
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(Original post by C0balt)
Hello
We were doing FP1 topic on calculus and the teacher was about to do an example using 1/x but he was like oops it's an exception and the integral of 1/x is actually ln x.

Is there any way for me to understand this thing having only done C1 and half of FP1 so far? Will I be able to understand it by the end of A2 FM?
I don't think it's too hard to understand this.

To say that \ln x is the integral of 1/x is equivalent to saying that:

a) \ln x differentiates to 1/x
c) the integral \int 1/x dx behaves just like the log function

The most important property of the log function is that \ln(ab) = \ln a + \ln b, so we want to show that we can make some integral \int 1/x dx have the same property (as well as the other properties of log, like \ln 1 = 0 and so on).

We can define a function L(x) as follows:

L(x) = \int_1^x \frac{1}{t} dt

which computes the area under 1/x from 1 to some arbitrary value x. This function is, in fact, the same as \ln x.

1. L(1) = 0 since the area under 1/x from 1 to 1 is 0

2. It should be more-or-less obvious that if we differentiate L(x) we get back to 1/t (since differentiating undoes integration)

3. L(ab) = L(a) + L(b) so it behaves like a log function

That's because L(ab) = \int_1^{ab} \frac{1}{t} dt = \int_1^a \frac{1}{t} dt + \int_a^{ab} \frac{1}{t} dt by splitting the area from 1 to ab into 2 chunks and adding them.

Now the first integral  \int_1^a \frac{1}{t} dt is simply L(a) by the defn of the function. Also if you make the substitution x=u/a you can show that the second integral \int_a^{ab} \frac{1}{t} dt is simply L(b) as we wanted.

So L(x) has all the properties of \ln x, and hence *is* \ln x and also (obviously?) differentiates to 1/x.

That is a more advanced way to answer the question but not inaccessible to a reasonably bright further maths student, I think.
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DFranklin
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(Original post by atsruser)
I don't think it's too hard to understand this.

To say that \ln x is the integral of 1/x is equivalent to saying that:

a) \ln x differentiates to 1/x
c) the integral \int 1/x dx behaves just like the log function
What is this "the log function you talk about?"

Explicitly, you mean log to base e, but what is this mysterious 'e' and where does it come from?

Everything else you say is correct, but also correct if you define log(x) = \alpha \int_1^x \frac{1}{t}\, dt for any alpha > 0.

At the end of the day, a lot depends on how you've defined e, I think.
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Prince Philip
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(Original post by DFranklin)
And I assume you prove the derivative of e^x by using the derivative of ln x and using similar reasoning...

[Obviously, the involvement of 'e' is a massive spanner in the works at A-level, since it's a completely unfamiliar number and the definitions of it are pretty intractable (\sum 1/n! is fairly easy to understand, but how the heck you work with the result is less so...)]
A Level starts by introducing e^x as a function with derivative e^x (no explanation but maybe an illustration using graphs of a^x functions and their gradient functions).

You can get by in life/applied maths without knowing much more.
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atsruser
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(Original post by DFranklin)
What is this "the log function you talk about?"

Explicitly, you mean log to base e, but what is this mysterious 'e' and where does it come from?

Everything else you say is correct, but also correct if you define log(x) = \alpha \int_1^x \frac{1}{t}\, dt for any alpha > 0.

At the end of the day, a lot depends on how you've defined e, I think.
You're right, I jumped the gun a bit by equating the L function to the natural log, explicitly.

However, my main point was to give someone a nudge towards seeing that there's a fairly obvious relationship between \int 1/x dx and log functions, and that by being aware that integration and differentiation are inverse operations to see that some log function should differentiate to 1/x.

As I'm sure you know, this can be made water-tight ("prove that there is a unique function L(x) with properties blah blah .. fundamental theorem of calculus blah blah .. unique inverse function E(L(x)) = x blah blah ..") but the details are out of place here.

(BTW, doesn't this discussion come round about every 2 months here, with the associated back-and-forth about chicken and egg definitions of e^x, e, ln(x), ooh that's a circular argument, no it isn't because I said such and such .. ?)
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davros
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(Original post by atsruser)
You're right, I jumped the gun a bit by equating the L function to the natural log, explicitly.

However, my main point was to give someone a nudge towards seeing that there's a fairly obvious relationship between \int 1/x dx and log functions, and that by being aware that integration and differentiation are inverse operations to see that some log function should differentiate to 1/x.

As I'm sure you know, this can be made water-tight ("prove that there is a unique function L(x) with properties blah blah .. fundamental theorem of calculus blah blah .. unique inverse function E(L(x)) = x blah blah ..") but the details are out of place here.

(BTW, doesn't this discussion come round about every 2 months here, with the associated back-and-forth about chicken and egg definitions of e^x, e, ln(x), ooh that's a circular argument, no it isn't because I said such and such .. ?)
I think "it" comes about in various guises quite regularly - there are often students who understandably want to see how things introduced at A level actually "work". This is perfectly laudable, but a lot of these students have an idealized view in which they think that A level deliberately "hides" the explanations away, and if they saw these then everything would be a lot easier. The reality is that setting up the foundations of things like calculus requires a serious amount of leg-work, and there are very good reasons why students should take things on trust at A level and concentrate on getting proficient in actually following the rules
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C0balt
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(Original post by notnek)
I'm not sure what kind of explanation you're looking for

E.g. are you happy that the integral of x^2 is x^3/3? What explanation has made you accept this?
I don't get why this weird thing called e and log come into play suddenly at 1/x


Yeah I guess so
I was told that integration is kind of inverse differentiation


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C0balt
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(Original post by tombayes)
Ok here is a 'basic i.e. non-formal' idea: the derivative of e^x is itself. i.e. let y=e^x then \frac{dy}{dx}=e^x=y
I'm lost here. I don't see why y=e^x is true in the first place x.x (as tenodthem says I'll learn in C3??)
If we rearrange  \frac{dy}{dx}=y we see that \frac{1}{y}dy=dx now integrating both sides gives \int \frac{1}{y}dy=x+c where c is constant but recall e^x=y \implies x=ln y
I get the last part but it's meaningless because I don't get the very first bit lol

so \int \frac{1}{y}dy=ln(y)+c



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Prince Philip
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(Original post by C0balt)
I don't get why this weird thing called e and log come into play suddenly at 1/x


Yeah I guess so
I was told that integration is kind of inverse differentiation


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You're right that integration is the reverse of differentiation. The integral of x^2 is x^3/3 because the derivative of x^3/3 is x^2.

But in the same way, can you look for something that differentiates to give 1/x? This is x^(-1) so using the same method as above, you may think that it's something involving x^0, but does this differentiate to give x^(-1)?

Since you haven't met logs or e yet, I'm not sure if you'll be able to get more of an explanation. I suggest either waiting until C3 or read a C3 textbook now / google it.
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davros
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(Original post by C0balt)
I don't get why this weird thing called e and log come into play suddenly at 1/x


Yeah I guess so
I was told that integration is kind of inverse differentiation


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Tbh, if you've only done C1 and FP1 then much of what's been written here isn't going to make a lot of sense to you taken out of context. You really need some background from C2-C4 and then revisit this issue
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TenOfThem
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(Original post by notnek)
A Level starts by introducing e^x as a function with derivative e^x (no explanation but maybe an illustration using graphs of a^x functions and their gradient functions).

You can get by in life/applied maths without knowing much more.
TBH this is the way I go - that is the definition of e that they get

Then, as has been said, the differential of ln(x) follows
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C0balt
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(Original post by atsruser)
I don't think it's too hard to understand this.

To say that \ln x is the integral of 1/x is equivalent to saying that:

a) \ln x differentiates to 1/x
c) the integral \int 1/x dx behaves just like the log function
I don't see why ln x differentiates to 1/x in the first place and I don't know what log function is (Good start isn't it....). We sort of touched on logarithm as a part of FP1 chapter called Linear Law but we winged it by just using mysterious law of logs x.x
The most important property of the log function is that \ln(ab) = \ln a + \ln b, so we want to show that we can make some integral \int 1/x dx have the same property (as well as the other properties of log, like \ln 1 = 0 and so on).
Ok I (kind of) am used to this log ab is log a+log b thing but uh...

We can define a function L(x) as follows:

L(x) = \int_1^x \frac{1}{t} dt

which computes the area under 1/x from 1 to some arbitrary value x. This function is, in fact, the same as \ln x.

1. L(1) = 0 since the area under 1/x from 1 to 1 is 0

2. It should be more-or-less obvious that if we differentiate L(x) we get back to 1/t (since differentiating undoes integration)

3. L(ab) = L(a) + L(b) so it behaves like a log function

That's because L(ab) = \int_1^{ab} \frac{1}{t} dt = \int_1^a \frac{1}{t} dt + \int_a^{ab} \frac{1}{t} dt by splitting the area from 1 to ab into 2 chunks and adding them.
Now the first integral  \int_1^a \frac{1}{t} dt is simply L(a) by the defn of the function. Also if you make the substitution x=u/a you can show that the second integral \int_a^{ab} \frac{1}{t} dt is simply L(b) as we wanted.

So L(x) has all the properties of \ln x, and hence *is* \ln x and also (obviously?) differentiates to 1/x.

That is a more advanced way to answer the question but not inaccessible to a reasonably bright further maths student, I think.
I see it but I don't at the same time because I don't get the first assumptions you made up there. It's like saying I know that fabulous is synonym to amazing but I don't know the definition of amazing or fabulous
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