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 February 24th, 2018, 02:57 AM #1 Member   Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 50 Thanks: 3 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/x-1)*(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: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond See here. (W|A). $$y=(A^x-B^x)^{1/x}$$ Take the natural logarithm of both sides: $$\ln(y)=\frac1x\ln(A^x-B^x)$$ Differentiate implicitly with respect to $x$: $$\frac{y'}{y}=-\frac{1}{x^2}\ln(A^x-B^x)+\frac1x\frac{\ln(A)A^x-\ln(B)B^x}{A^x-B^x}$$ 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: 50 Thanks: 3 Math Focus: Number theory, Applied maths Thanks Thank you for such a quick reply Thanks from greg1313

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