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

1. May 20, 2003

MathematicalPhysicist

Last edited by a moderator: Apr 20, 2017
2. May 20, 2003

Lonewolf

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?

3. May 20, 2003

Mulder

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

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

4. May 20, 2003

newton1

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??!!)

5. May 20, 2003

newton1

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

6. May 20, 2003

Lonewolf

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

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

7. May 20, 2003

FZ+

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

8. May 20, 2003

Tom Mattson

Staff Emeritus
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!

9. May 20, 2003

Lonewolf

Shouldn't that be

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

10. May 20, 2003

Tom Mattson

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

11. May 21, 2003

plus

just differentiate ln|f(x)|.

12. May 21, 2003

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

Last edited by a moderator: May 21, 2003
13. May 21, 2003

Lonewolf

f'(x) = -1/x2

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

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

14. May 21, 2003

LogicalAtheist

For the expression:

∫X

15. May 21, 2003

Lonewolf

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.

16. May 21, 2003

KLscilevothma

another mistake

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

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

17. May 21, 2003

Lonewolf

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

18. May 21, 2003

Thoth

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)|.

19. May 21, 2003

Lonewolf

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

20. May 21, 2003

Thoth

Please read my original post carefully, I said, “In general: if f (x) = cx^n then…”
I didn’t have to, n can be anything, and however, n is not -n. I did not say if f (x) = cx -n
Exactly!! I think you got it now.