Is This Proof Valid for Continuous Partial Derivatives?

In summary, if u : R^2 \to R has continuous partial derivatives at a point (x_0,y_0) then 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.
  • #1
Mandark
42
0
If [tex]u : R^2 \to R[/tex] has continuous partial derivatives at a point [tex](x_0,y_0)[/tex] show that:

[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
 
Physics news on Phys.org
  • #2
I'm not sure if this is the cause of your problem, but you seem to be disregarding the fact that [itex]\epsilon_1 \equiv \epsilon_1(y_0 + \Delta y), \epsilon_2 \equiv \epsilon_2(x_0 + \Delta x)[/itex]. It doesn't follow from the fact that for every Delta y, epsilon 1 goes to zero if Delta x goes to zero that epsilon 1 goes to zero if we approach the point from any direction.
 
  • #3
hi! could you please clear this doubt...
there is a theorem stating that if the first partial derivatives are continuous, then the function in 2 variables is differentiable. Is it enough to prove this in case of a split function?
 

FAQ: Is This Proof Valid for Continuous Partial Derivatives?

What is a continuous partial derivative?

A continuous partial derivative is a mathematical concept used to describe the rate of change of one variable in a function with respect to another variable while holding all other variables constant. It is a type of derivative that is defined for multi-variable functions.

How is a continuous partial derivative calculated?

A continuous partial derivative is calculated by taking the ordinary derivative of a multi-variable function with respect to one of its variables while holding all other variables constant. This can be done using the standard rules of calculus, such as the power rule, product rule, and chain rule.

Why is the concept of continuous partial derivative important?

The concept of continuous partial derivative is important because it allows us to understand and analyze the behavior of multi-variable functions. It is used in many fields of science and engineering, such as physics, economics, and engineering, to model and solve real-world problems.

What is the difference between a continuous partial derivative and a regular derivative?

The main difference between a continuous partial derivative and a regular derivative is that a regular derivative is calculated with respect to only one variable, while a continuous partial derivative is calculated with respect to one variable while holding all other variables constant. Additionally, a regular derivative is defined for single-variable functions, while a continuous partial derivative is defined for multi-variable functions.

Can a continuous partial derivative be negative?

Yes, a continuous partial derivative can be negative. This means that the function is decreasing with respect to the variable being differentiated, while holding all other variables constant. A negative continuous partial derivative is often interpreted as a negative slope or a decreasing rate of change.

Back
Top