\fExample 1 Show that f(x, y) = x2 + y is differentiable at (1, 0). Solution f(x, y) is differentiable at (a, b) = (1,0) if | R1, (1,0) (2, y) | lim = 0 (x,y) -(1,0) |1(x, 3) - (1,0)11 Find We are given that f(x, y) = a2 + y2. R1, (1,0) (x , y) = f(x, y) - L(1,0) (x, y) = x2 + y2 - (1+ 2(x -1)) We know that = (a -1)2+ 32 L(1,0) (x, y) = f(1, 0) + fx(1, 0)(x - 1) + fy (1,0) (y - 0) . f(1,0) = 1 . fx(x, y) = 2x, so fx (1, 0) = 2 . fy(x, y) = 2y, so fy (1, 0) = 0 Hence L(1,0) (x, y) = 1+ 2(x - 1)ac2 y Determine whether f(x, y) = 2 2 +2 2 if ( x, y) * (0, 0) is differentiable at (0, 0) 0 if ( x , y) = (0,0) Solution | R1, (0,0) (2, y)I For f to be differentiable at (0, 0), we need lim = 0. (z,y) -(0,0) Va2 + 32 If we approach the limit along y = x, we get R 1, (0,0) (2, y) | [ R1, (0,0) (20, 2C) | lim = lim = lim (a,y) -(0,0) Vx2+ y2 x-+0 -+0 (ac 2 + 20 2 ) 3/2 = lim (2ac2 ) 3/2 = lim x-+0 23/2123 1 = lim x-+0 93/2 1 = 23/2 Therefore, the limit cannot equal 0, and hence f is not differentiable at (0, 0)