# Proof f'(x)/f(x)=|f(x)|

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I may be being naive here, but I can't see how that's true...
Consider f(x)=x. Then f'(x)=1

f'(x)/f(x)=1/x which is not equal to |x|.

Am I right in thinking this?

It should probably say [inte](f'(x)/f(x))dx = ln|f(x)|.

Try integrating cos(x) / sin(x) for example.

i never saw this equation before
if f'(x) is a derivative of f(x)...
this equation should be wrong
if not, then what is it??(i mean maybe have other meaning??!!)

Originally posted by Mulder
It should probably say [inte](f'(x)/f(x))dx = ln|f(x)|.

Try integrating cos(x) / sin(x) for example.

[inte](f'(x)/f(x))dx = ln|f(x)|.
yes....i think this is the right answer

I wouldn't use that site much more...

&int;1/x2dx= tanh-1x+c?!

Seems to be a straightforward case of integration by substitution, doesn't it? Which is derived from chain rule...

Tom Mattson
Staff Emeritus
Gold Member
It's also wrong, which is Lonewolf's point. It should be:

&int;1/(x2+1)dx= tanh-1x+c

Funny thing is, they got the derivative of tanh-1(x) right!

Shouldn't that be

&int;1/(1-x2)dx= arctanh x + c?

Tom Mattson
Staff Emeritus
Gold Member
Oops--you're right. I got tanh-1(x) confused with tan-1(x).

Originally posted by newton1
[inte](f'(x)/f(x))dx = ln|f(x)|.
yes....i think this is the right answer

just differentiate ln|f(x)|.

Thoth
This is actually f’ (x)/f (x)=|f’ (x)/f (x)|. Let’s make some examples:

If f (x)=x then f’ (x)=1, therefore, f’ (x)/f (x)=1/x and |f’ (x)/f (x)|=Square-root (1/x)^2=1/x

If f (x)=x^2 then f’ (x)=2x, therefore, f’ (x)/f (x)=2/x and |f’ (x)/f (x)|= Square-root (2/x)^2=2/x

How about if f (x)= -x then f’ (x) = -1, therefore, f’ (x)/f (x)= -1/-x=1/x and again |f’ (x)/f (x)| =1/x
Obviously f’ (x) / f (x) is always equal to |f’ (x)/f (x)|

In general: if f (x) = cx^n then f’ (x) = ncx^(n-1) so f’ (x)/f (x) = ncx^n/cx^(n+1)= nx^n/x^(n+1) =nx^n/(x^n)*x=n/x

So f’ (x)/f (x) for x^3 is 3/x or x^4 is 4/x and so on (always n/x). This equation states that there is always a singularity at x and it exists at first and third quadrants. I hope this was helpful

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f'(x) = -1/x2

f'(x)/f(x) = -1/x

This is not equal to |f'(x)/f(x)|

LogicalAtheist
For the expression:

∫X

It's an integral sign. Integration is a part of calculus where we find areas of regions, naively speaking. There's no real general method of performing integration, but there are certain standard integrals. The site listed above gives some examples of standard integrals, although a few of them are wrong. I'm sure someone can give a more precise definition than I have.

another mistake

[inte] tan hx
=/= ln|cos hx| + C
= ln|sec hx| +C
(another mistake from the website)

What does the "[inte]" mean
For example, a function f(x) = x^2
The first derivative of f(x), f'(x) = 2x

Now we are given f'(x) = 2x , how can we find f(x)? We use integration.
[inte]2xdx = x^2 + C (where C is the constant of integration)

Defination: Let f(x) be defined on [a,b]
If F'(x)=f(x), then F is called the primitive function of f and we write it as F(x)=[inte]f(x)dx.
Since F(x)+C is also a primitive function of f, therefore [inte]f(x)dx= F(x) + C where C is the constant of integration

Not all functions have a closed form integral. One such famous one is the probability density function of the Gaussian distribution

Thoth
Originally posted by Lonewolf

f'(x) = -1/x2

f'(x)/f(x) = -1/x

This is not equal to |f'(x)/f(x)|

1/x or better said x-1 is not in the form of cxn, but it is in the form of cx(-n), which generally exists in the second and forth quadrants with equation of –n/x. In this case we would have -|f’ (x) / f (x)|.

You didn't say what n had to be. f(x) represents a general function. Your 'proof' doesn't cover a general function, only polynomials.
Try f(x) = sin(x) for size

Thoth
Originally posted by Lonewolf
f(x) represents a general function.
Please read my original post carefully, I said, “In general: if f (x) = cx^n then…”
Originally posted by Lonewolf
You didn't say what n had to be.
I didn’t have to, n can be anything, and however, n is not -n. I did not say if f (x) = cx -n
Your 'proof' doesn't cover a general function, only polynomials.
Exactly!! I think you got it now.

I didn’t have to, n can be anything

I'm afraid you do, it makes a whole lot of difference. Mathematical proofs require precision and rigour. n could have been included in the whole set of real numbers, which does include negative numbers, or even the set of complex numbers, which leads to all hell breaking loose.

Your theorem was f'(x)/f(x) = |f'(x)/f(x)|. You should have specified earlier that f was a polynomial function.

I'd go with earlier posts and say that it should have been

&int;f'(x)/f(x) dx = ln|f(x)| + C

Hmm, furthermore, it's not true.

Consider f(x) = x, f'(x) = 1

f'(x)/f(x) = 1/x

Now, let x = -1

f'(-1)/f(-1) = 1/-1 = -1
|f'(-1)/f(-1)| = |1/-1| = |-1| = 1

1 is not equal to -1

Thoth
Lonewolf

Please read my original post once again, I am sure the problem is not you but my English, since English is not my first or second or even third language. I apologize if I do not know the English languages as well as you do (after all this is probably your language from birth.) Perhaps instead of the middle I should have started the discussion with Polynomials at the top. However, I am certain that those who are looking for excuses to argue, would have find something else to pick on no matter what.

The other problem is that we might both have to go and look at the math 101 on mathematical symbols and denotations. For example, to show a general polynomial equation, c is usually means a Constant, n denotes a positive number. To signify a negative number, mostly -n is used. To indicate a complex number, i and sometimes u is used and never n.

About if x=-1, please read my original post line by line. Look where I mentioned the first and third quadrant for f’ (x)/f (x)=|f’ (x)/f (x) | to be true. In the first quadrant x cannot be a negative number and in the third quadrant where x can be negative, f (x) also have to be negative. Therefore, f (-x)=-f (x). This means that in the third quadrant f (-x)=-x and not x, hence, f’ (-1)=-1 and not 1 as you mistakenly stated.

So f’ (-1) / f (-1) = -1/-1=1 which is also |f’ (-1) / f (-1)|=1

Furthermore, I never did or ever will claim that I am proofing anything since I neither have time nor desire to proof mathematics in a forum such as this. I only was trying to help someone about an equation in a faulty Internet site. This is a forum for discussion and exchange of ideas and not a science symposiums or mathematical proofing society.
Take it easy life is short

The other problem is that we might both have to go and look at the math 101 on mathematical symbols and denotations.

I never knew there was such a class. To paraphrase Lewis Carroll, 'my words mean whatever I want them to mean'. This is why we need to define the terms we use. Different people have different conventions. For example, engineers like to use j to represent &radic;(-1).

Think about the usefulness of your function, and the usefulness of the proposed integral, &int;f'(x)/f(x)dx=ln|f(x)|+C. This occurs frequently in calculus, and is applicable in general to any real function where f(x) is never 0, and applicable to any function from the reals minus the points where the function is 0 to the reals for all x.

I appreciate that you were trying to help. I'm just making a point.

Thoth
Yes, the inquisitive Alice in wonderland. I actually appreciate your inclination to investigate intrusively, and I value your friction with me. Where we would be if there were no abrasions in life. About the degree of usefulness of a proposition, this isn’t really the point. The question is, which expression mostly resembles this wrong expression: f’ (x)/ f (x) = |f (x)|.

About the frequency of occurrences in calculus, take a look at the following and tell me if this not the most common questions in calculus and pre calculus.

The function F (x) is as follows:

F (x) = x where x>0
F (x) = undefined where x=0
F (x) = -x where x<0
What is f ‘(x)/f (x)? Or what is f’ (x) at x=-1 for example.

Most of the people make the same mistake that you did and intuitively think that f’ (-1) = 1. However, if you release yourself from the bondage of intuition and stuck with the art of logic, you immediately realize the trick (which teachers and books love to do.)

This is obviously not continues function and there is a singularity and un-continuity in F (x). As there is with any f (x) = cx n . This was of course all discussed in my initial writing regarding this issue. However, perhaps the way I organize it wasn’t quite clear by not being written in a more customary format for pre calculus students.