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 February 24th, 2018, 02:57 AM #1 Member   Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 57 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: 7,968 Thanks: 1152 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: 57 Thanks: 3 Math Focus: Number theory, Applied maths Thanks Thank you for such a quick reply Thanks from greg1313 Tags differentiation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MathDino Pre-Calculus 2 April 14th, 2015 08:58 AM Mr Davis 97 Calculus 1 March 9th, 2015 04:50 AM jiasyuen Calculus 5 January 3rd, 2015 09:31 AM jiasyuen Calculus 2 December 28th, 2014 06:24 AM jiasyuen Calculus 3 December 27th, 2014 06:01 PM

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