User Name Remember Me? Password

 Calculus Calculus Math Forum

 May 19th, 2010, 02:18 AM #1 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Partial derivatives Hi, I have a question: -- Let f(x,y) = x^2/x+y Obtain the first two partial derivatives f_x and f_y and evaluate them at point (2,2). -- Could someone please confirm my answer? -- f_x; make y constant. = 2x(x + y)^-1 + x^2(-x + y)^-2 f_x = 2x / (x + y) + x^2 / (-x + y)^2 If we sub (2,2) we have: 2(2) / (2 + 2) + 2^2 / (-2 + 2)^2 = 1 + 0 = 1 ANS -- f_y; make x constant. = 0(x + y)^-1 + x^2(-x + y)^-2 f_y = x^2 / (-x + y)^2 If we sub (2,2) we have: 2(2) / (-2 + 2)^2 = 0 ANS -- .. ? wulfgarpro. May 19th, 2010, 02:30 AM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Partial derivatives With y constant d/dx x^2(x + y)^-1 = 2x(x + y)^-1 - x^2(x + y)^-2 With x constant d/dx x^2(x + y)^-1 = -x^2(x + y)^-2 May 20th, 2010, 02:54 AM #3 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Re: Partial derivatives Evaluated at point (2,2), I get: f_x = 12/16 = 3/4 f_y = -4/16 = -1/4 ? wulfgarpro. May 20th, 2010, 02:58 AM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Partial derivatives Correct. Tags derivatives, partial 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 johngalt47 Calculus 2 November 23rd, 2013 11:44 AM r-soy Calculus 7 September 25th, 2012 01:43 AM greg1313 Calculus 3 July 7th, 2010 01:09 PM remeday86 Calculus 3 March 28th, 2009 08:25 AM remeday86 Calculus 3 March 28th, 2009 06:50 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      