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 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.

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