
Math General Math Forum  For general math related discussion and news 
 LinkBack  Thread Tools  Display Modes 
February 24th, 2018, 02:57 AM  #1 
Member Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 47 Thanks: 2 Math Focus: Number theory, Applied maths  differentiation
What is the differential of y =(A^x B^x)^(1/x) ? I make it dy/dx = (1/x)*(A^x  B^x)^(1/x1)*(A^x B^x)*LOGe(10) = (1/x)*(A^x  B^x)^(1/x)*LOGe(10) but trials show this is wrong, because y change for x change by 1 is less than this increasingly less as x increases. Where am I going wrong and what is the correct differential? 
February 24th, 2018, 03:04 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,764 Thanks: 1012 Math Focus: Elementary mathematics and beyond 
See here. (WA). $$y=(A^xB^x)^{1/x}$$ Take the natural logarithm of both sides: $$\ln(y)=\frac1x\ln(A^xB^x)$$ Differentiate implicitly with respect to $x$: $$\frac{y'}{y}=\frac{1}{x^2}\ln(A^xB^x)+\frac1x\frac{\ln(A)A^x\ln(B)B^x}{A^xB^x}$$ I'll let you finish up. Last edited by greg1313; February 24th, 2018 at 05:46 AM. 
February 24th, 2018, 03:24 AM  #3 
Member Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 47 Thanks: 2 Math Focus: Number theory, Applied maths  Thanks
Thank you for such a quick reply 

Tags 
differentiation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Differentiation  MathDino  PreCalculus  2  April 14th, 2015 08:58 AM 
Difference between vector differentiation and parametric differentiation  Mr Davis 97  Calculus  1  March 9th, 2015 04:50 AM 
Differentiation  jiasyuen  Calculus  5  January 3rd, 2015 09:31 AM 
Differentiation  jiasyuen  Calculus  2  December 28th, 2014 06:24 AM 
Differentiation  jiasyuen  Calculus  3  December 27th, 2014 06:01 PM 