If I have some function like y = kx/(1+x)

 

How do I take the derivative of dln(y)/dln(x)

 

ln is natural log...

 

Can someone do like a step by step kind of a thing...

What have you tried so far? Chain rule? Quotient rule?

well it could work if you'd either of the rules depending on how you think about it algebraically. I just don't know what to do with those logs in the derivative portion.

 

I know how to apply dy/dx and find the derivative but what do you do when it calls for dlny/dlnx.

 

what does that mean?

Have you tried substitution? Say, u=lnx ?

 

What attempts have you made so far?

So this is not a homework problem and I am not an undergrad... math is not my thing but I use it here and there... I just want to understand the significance of taking dlny/dlnx of a function... and how that works...

First up, partial fractions to make this vaguely approachable.

[math]y=\frac{k x}{1+x}=k-\frac{k}{1-x}=k\left( 1-\frac{1}{1-x}\right)[/math]

Then some substitution: take [imath]u:=\ln(y)[/imath] and [imath]v:=\ln(x)[/imath].

[math]e^{u} = k\left( 1-\frac{1}{1-e^v}\right)[/math]

Take logs.

[math]u = \ln(k) + \ln(1 - \tfrac{1}{1-e^v} )[/math]

Finally, differentiate.

[math]\frac{du}{dv} = \frac{ \tfrac{d}{dv} \left( 1-\tfrac{1}{1-e^v} \right)}{1-\tfrac{1}{1-e^v}}=- \frac{ \tfrac{e^v}{(1-e^v)^2} }{1-\tfrac{1}{1-e^v}}[/math]

Throw [imath]x[/imath] back in and simplify as much as you feel like...

[math]\frac{du}{dv} = \frac{-x}{(1-x)^2 (1-\tfrac{1}{1-x})}= \frac{-x}{x^2 - x}= \frac{-1}{x-1}[/math]

And voilà!

[math]\frac{d\ln(y)}{d\ln(x)}= \frac{-1}{x-1}[/math]

And no, I have no idea why you would want to do that, or why it would be relevant to anything.

Edited January 17, 201015 yr by the tree
sign error

I do have something like an idea: http://www.scienceforums.net/forum/showthread.php?t=47739 . You made a sign error in the denominator when taking the derivative d/dv (1/(1-exp(v)), btw.

You made a sign error in the denominator when taking the derivative d/dv (1/(1-exp(v)), btw.

So I did. Fixed and now the answer makes a lot more sense, thanks.

Where did the k go?

 

Is the answer not y = k - 1/(1+x) after taking those derivatives ?

How much does k vary with x or ln(k) with v ?

Where did the k go?

It disappears around the point that you take it out as a common factor.

[imath]e^{u} = k\left( 1-\frac{1}{1-e^v}\right)[/imath]...Take logs....[imath]u = \ln(k) + \ln(1 - \tfrac{1}{1-e^v} )[/imath]

Then [imath]\frac{\mbox{d}u}{\mbox{d}v}= \frac{\mbox{d}}{\mbox{d}v} \ln(k) + \frac{\mbox{d}}{\mbox{d}v} \ln(1 - \tfrac{1}{1-e^v} )[/imath]

And obviously: [imath] \tfrac{\mbox{d}}{\mbox{d}v} \ln(k)=0[/imath]

First up, partial fractions to make this vaguely approachable.

[math]y=\frac{k x}{1+x}=k-\frac{k}{1-x}=k \left( 1-\frac{1}{1-x} \right)[/math]

 

Shouldn't this be [math]y=\frac{k x}{1+x}=k-\frac{k}{1+x}=k \left( 1-\frac{1}{1+x} \right)[/math]? Which means that the correct answer would be [math]\frac {x}{x+1}[/math].

Edited January 9, 201114 yr by Shadow

Also, I believe the process would be a lot less messy if you rewrote

 

[math]\frac{\mbox{d} \ln(\frac{kx}{x+1})}{\mbox{d} \ln(x)}[/math]

 

as

 

[math]\frac{\mbox{d} \ln(\frac{kx}{x+1})}{\mbox{d }x} \cdot \frac{1}{\frac{\mbox{d} \ln(x)}{\mbox{d} x}}[/math]

Yes and yes, although I did do that a year ago so I think it's too late to be overly bothered about it.

The correct answer to taking the derivative [math]

 

\frac{{d\ln y}}{{d\ln x}}

 

[/math] of [math]

 

y = \frac{{kx}}{{1 + x}}

 

[/math]

 

is

 

[math]

\frac{{d\ln y}}{{d\ln x}} = \frac{x}{y}\frac{{dy}}{{dx}} = \frac{x}{y}\left( {\frac{k}{{1 + x}} - \frac{{kx}}{{(1 + x)^2 }}} \right) = \frac{{(1 + x)x}}{{kx}}\left( {\frac{k}{{1 + x}} - \frac{{kx}}{{(1 + x)^2 }}} \right) = \frac{{1 + x}}{k}\left( {\frac{k}{{1 + x}} - \frac{{kx}}{{(1 + x)^2 }}} \right) = 1 - \frac{x}{{1 + x}}

 

[/math]

Edited January 12, 201114 yr by apricimo

$\frac{dln(y)}{dln(x)}=\frac{x}{y}\frac{dy}{dx}$