If [tex]u : R^2 \to R[/tex] has continuous partial derivatives at a point [tex](x_0,y_0)[/tex] show that:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]u(x_0+\Delta x, y_0+\Delta y) = u_x(x_0,y_0) + u_y(x_0,y_0) + \epsilon_1 \Delta x + \epsilon_2 \Delta y[/tex], with [tex]\epsilon_1,\, \epsilon_2 \to 0[/tex] as [tex]\Delta x,\, \Delta y \to 0[/tex]

I know this can be proved using MVT, but I tried to prove this another way only my proof doesn't use the continuity of both partial derivatives so I thought there'd be an error but I couldn't spot it, so I was hoping somebody else could. Here is my proof:

I will use the result that for a differentiable function f,

[tex]f(x+h) = f(x) + f'(x) h + \epsilon h[/tex] where [tex]\epsilon[/tex] is a function of h and goes to 0 as h goes to zero. (Follows from the definition of the derivative.)

[tex]u(x_0+\Delta x,y_0+\Delta y) - u(x_0,y_0)[/tex]

[tex]= u(x_0,y_0+\Delta y) + u_x(x_0,y_0+\Delta y) \Delta x + \epsilon_1 \Delta x - u(x_0,y_0)[/tex]

[tex]= (u(x_0,y_0) + u_y(x_0,y_0) \Delta y + \epsilon_2\Delta y) + u_x(x_0,y_0+\Delta y) \Delta x + \epsilon_1\Delta x - u(x_0,y_0)[/tex]

[tex]= u_y(x_0,y_0) \Delta y + u_x(x_0,y_0+\Delta y) \Delta x + \epsilon_1 \Delta x + \epsilon_2 \Delta y[/tex]

Here [tex]\epsilon_1,\, \epsilon_2 \to 0[/tex] as [tex]\Delta x,\, \Delta y \to 0[/tex].

I will be done if I can prove [tex]u_x(x_0, y_0 + \Delta y) \Delta x = u_x(x_0, y_0)\Delta x + \epsilon_3 \Delta x[/tex] with [tex]\epsilon_3 \to 0[/tex] as [tex]\Delta x,\, \Delta y \to 0[/tex], but this follows from the continuity of u_x at the point [tex](x_0, y_0)[/tex].

Thanks

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# Continuous Partial Derivative

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