Derivative of Natural Log - How Does 'a' Affect the Derivative of Ln(ax)?

In summary, the conversation discusses the derivative and integral of a function of the form Ln(ax), with the question of whether the derivative is always 1/x regardless of the initial value of 'a'. It is clarified that while the derivative is 1/x, the integral is not Ln(3x) due to the arbitrary constant 'C'. The conversation then delves into understanding this concept through graphing and observing the reciprocal relationship between the graphs of ln(x) and ex. There is a brief discussion on the use of the terms 'reciprocal' and 'inverse', with the conclusion that graphing can provide a better understanding of this concept.
  • #1
grscott_2000
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0
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?
 
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  • #2
grscott_2000 said:
but the integral of 1/x is not Ln(3x).

Are you sure?

Don't forget about the arbitary constant that's associated with an indefinite integral.
 
  • #3
ln(3x)= ln(x)+ ln(3). As George Jones said (love his singing!), "don't forget about the arbitrary constant".
 
  • #4
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.
 
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  • #5
Schrodinger's Dog said:
...
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 reciprocal relationship to each other...

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

Thanx. :)
 
  • #6
VietDao29 said:
Err, not pretty sure what you mean by "reciprocal relationship"? :\ Can you explain a little bit more?

Thanx. :)

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:-

[itex]\frac{3}{1}\times\frac{1}{3}=1\rightarrow3=1/3 [/itex]

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..
 
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  • #7
Schrodinger's Dog said:
Oh yeah reciprocals aren't studied here any more although you may see the word used.

the reciprocal is:-

[itex]3\times\frac{1}{3}=1\rightarrow3=1/3 [/itex]

in other words a relationship which equates one value with 1.

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
Schrodinger's Dog said:
The inverse of ex is ln(ex)

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

A reciprocal is an inverse, that's kind of what I was saying rather poorly though obviously :smile: Why this is is best observed by looking at a graph of the function ln and ex and observing pictorially the rates of change or the area under the graph when different values are used, to note the arbitrary nature of constant C. I hope that makes more sense.

Err, not sure if I understand this correctly, but, are you implying that reciprocal is the same as inverse? No, it's not. [tex]\frac{1}{e ^ x} \neq \ln (x)[/tex]
 
  • #9
VietDao29 said:
No, it's not. The inverse of ex is ln(x), not ln(ex) (which is just x).

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

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

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 :smile:
 
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  • #10
Schrodinger's Dog said:
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.

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

FAQ: Derivative of Natural Log - How Does 'a' Affect the Derivative of Ln(ax)?

What is the derivative of ln(x)?

The derivative of ln(x) is 1/x.

What is the general rule for finding the derivative of a natural log?

The general rule for finding the derivative of a natural log is to use the formula d/dx ln(x) = 1/x.

How do you find the derivative of ln(u), where u is a function of x?

To find the derivative of ln(u), where u is a function of x, you can use the chain rule: d/dx ln(u) = (1/u)*(du/dx).

Can the derivative of natural log be negative?

Yes, the derivative of natural log can be negative. This occurs when the value of x is less than 1.

How is the derivative of natural log used in real life applications?

The derivative of natural log is used in many real life applications, such as in finance and economics to calculate continuous growth rates, in physics to model exponential decay, and in biology to measure population growth rates. It is also used in calculus to solve optimization problems and find maximum and minimum values.

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