apricimo Posted January 14, 2010 Share Posted January 14, 2010 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... Link to comment Share on other sites More sharing options...
ydoaPs Posted January 14, 2010 Share Posted January 14, 2010 What have you tried so far? Chain rule? Quotient rule? Link to comment Share on other sites More sharing options...
apricimo Posted January 14, 2010 Author Share Posted January 14, 2010 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? Link to comment Share on other sites More sharing options...
ydoaPs Posted January 14, 2010 Share Posted January 14, 2010 Have you tried substitution? Say, u=lnx ? What attempts have you made so far? Link to comment Share on other sites More sharing options...
apricimo Posted January 14, 2010 Author Share Posted January 14, 2010 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... Link to comment Share on other sites More sharing options...
the tree Posted January 17, 2010 Share Posted January 17, 2010 (edited) 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, 2010 by the tree sign error 1 Link to comment Share on other sites More sharing options...
timo Posted January 17, 2010 Share Posted January 17, 2010 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. Link to comment Share on other sites More sharing options...
the tree Posted January 17, 2010 Share Posted January 17, 2010 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. Link to comment Share on other sites More sharing options...
apricimo Posted January 6, 2011 Author Share Posted January 6, 2011 Where did the k go? Is the answer not y = k - 1/(1+x) after taking those derivatives ? Link to comment Share on other sites More sharing options...
imatfaal Posted January 7, 2011 Share Posted January 7, 2011 How much does k vary with x or ln(k) with v ? Link to comment Share on other sites More sharing options...
the tree Posted January 8, 2011 Share Posted January 8, 2011 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] Link to comment Share on other sites More sharing options...
Shadow Posted January 9, 2011 Share Posted January 9, 2011 (edited) 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, 2011 by Shadow Link to comment Share on other sites More sharing options...
Shadow Posted January 10, 2011 Share Posted January 10, 2011 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] Link to comment Share on other sites More sharing options...
the tree Posted January 10, 2011 Share Posted January 10, 2011 Yes and yes, although I did do that a year ago so I think it's too late to be overly bothered about it. Link to comment Share on other sites More sharing options...
apricimo Posted January 12, 2011 Author Share Posted January 12, 2011 (edited) 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, 2011 by apricimo Link to comment Share on other sites More sharing options...
Stefanorgc Posted March 29, 2018 Share Posted March 29, 2018 $\frac{dln(y)}{dln(x)}=\frac{x}{y}\frac{dy}{dx}$ Link to comment Share on other sites More sharing options...
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