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

[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?

 

[George Jones]

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.

 

[HallsofIvy]

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

 

[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.

 

[VietDao29]

...
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. :)

 

[Schrodinger's Dog]

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..

 

[cristo]

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!

 

[VietDao29]

The inverse of e^x is ln(e^x)

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

A reciprocal is an inverse, that's kind of what I was saying rather poorly though obviously Why this is is best observed by looking at a graph of the function ln and e^x 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]

 

[Schrodinger's Dog]

No, it's not. The inverse of e^x is ln(x), not ln(e^x) (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

 

[VietDao29]

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.