# Derivative of Natural Log

1. Jul 4, 2007

### grscott_2000

If we have a function of the form

Ln(ax)

Is it the case that the derivative is simply 1/x no matter what the initial value of 'a' might be? Or do we take into account 'a' in some way

The thing i need clarifying is this.... If we have Ln(3x), the derivative is 1/x, but the integral of 1/x is not Ln(3x). How can this be so?

2. Jul 4, 2007

### George Jones

Staff Emeritus
Are you sure?

Don't forget about the arbitary constant that's associated with an indefinite integral.

3. Jul 4, 2007

### HallsofIvy

Staff Emeritus
ln(3x)= ln(x)+ ln(3). As George Jones said (love his singing!), "don't forget about the arbitrary constant".

4. Jul 5, 2007

### Schrodinger's Dog

Using the chain rule will show an answer and is in fact how the ln function rule is derived I think.

I'm not sure it will be much help though but what maybe is graphing several values of the integral and differentials of log and exp and looking at their inverse relationship to each other.

As said the arbitrary constant is just that. It does not make more than an arbitrary contribution to the function and thus can be accounted for by C in the Integral.

Last edited: Jul 5, 2007
5. Jul 5, 2007

### VietDao29

Err, not pretty sure what you mean by "reciprocal relationship"? :\ Can you explain a little bit more?

Thanx. :)

6. Jul 5, 2007

### Schrodinger's Dog

Oh yeah reciprocals aren't studied here any more either as such although you may see the word used in passing.

the reciprocal of 3 is:-

$\frac{3}{1}\times\frac{1}{3}=1\rightarrow3=1/3$

From wiki.

In mathematics, the multiplicative inverse of a number x, denoted 1/x or x−1, is the number which, when multiplied by x, yields 1..

Last edited: Jul 5, 2007
7. Jul 5, 2007

### cristo

Staff Emeritus
I get what a reciprocal is, but I'm not sure what you mean by the reciprocal relationships between the graphs of the integrals and derivatives of logx and expx. I guess that this is what VietDao is questioning too. Could you clarify?

Welcome back by the way SD. Not seen you around here for ages!

8. Jul 5, 2007

### VietDao29

No, it's not. The inverse of ex is ln(x), not ln(ex) (which is just x).

Err, not sure if I understand this correctly, but, are you implying that reciprocal is the same as inverse? No, it's not. $$\frac{1}{e ^ x} \neq \ln (x)$$

9. Jul 5, 2007

### Schrodinger's Dog

Oops yeah it's been a while since I did this

I'm willing to accept I used the wrong term, but apart from the misuse of language in that particular word, the advice is solid, no? Graphing always help me to take a visualisation and make it concrete.

Ammended and elided all posts to make it less wrong

Last edited: Jul 5, 2007
10. Jul 5, 2007

### VietDao29

Yup, I have no objection for the advantages of graphing. It does help a ton in visualization.