Differentiable Vector Functions

In summary, the function is differentiable at (1,1) because the partial derivatives exist and are continuous.
  • #1
Moridin
692
3

Homework Statement



Show through direct application of definition that the function

[tex]f(x,y) = xy[/tex]

is differentiable at (1,1)

The Attempt at a Solution



I know that all functions of the class C1 are differentiable and that a function is of the class C1 if its partial derivatives exists and are continuous.

So do I prove that the two partial derivatives exist by showing that the following to limits exists?

[tex]\frac{\partial f}{\partial x}(a,b) = \lim_{h \to 0} \frac{f(a+h,b) - f(a,b)}{h}[/tex]

[tex]\frac{\partial f}{\partial y}(a,b) = \lim_{k \to 0} \frac{f(a,b+k) - f(a,b)}{k}[/tex]

The, the following obtains:

[tex]\frac{\partial f}{\partial x}(1,1) = \lim_{h \to 0} \frac{f(1+h,1) - f(1,1)}{h} = \lim_{h \to 0} \frac{(1+h)1 - 1}{h} = \lim_{h \to 0}\frac{h}{h} = 1[/tex]

[tex]\frac{\partial f}{\partial y}(1,1) = \lim_{k \to 0} \frac{f(1,1+k) - f(1,1)}{k} = \lim_{k \to 0} \frac{(1(1+k) - 1}{k} = \lim_{k \to 0}\frac{k}{k} = 1[/tex]

Since both partial derivatives are elementary functions, they are continuous, correct? If so, this ought to show that the function is differentiable in (1,1)?

Thank you for your time. Have a nice day.
 
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  • #2
Yes, given that you have the theorem "if the partial derivatives of function f exist and are continuous at a point then the function is differentiable there" then what showing the partial derivatives are continuous would do it. However, I don't see that you have proven the partial derivatives are continuous. You have only calculated the partial derivatives at (1, 1). That does not show that they are elementary functions in a neighborhood of (1, 1) or that they are continuous.

One could also show it, without that theorem, directly from the definition of "differentiable".
 
  • #3
HallsofIvy said:
Yes, given that you have the theorem "if the partial derivatives of function f exist and are continuous at a point then the function is differentiable there" then what showing the partial derivatives are continuous would do it. However, I don't see that you have proven the partial derivatives are continuous. You have only calculated the partial derivatives at (1, 1). That does not show that they are elementary functions in a neighborhood of (1, 1) or that they are continuous.

So I have shown that the partial derivatives exist at the point (1,1). Can I then make the following argument:

1. If the functions have partial derivatives at all, they must be [insert the two partial derivatives here] (by taking the two partial derivative of the function with the short-hand rules)
2. If the partial derivatives exist, they are elementary functions and thus continuous. (by demonstration)
3. The partial derivatives exists at the point (1,1)
4. The partial derivatives are continuous at point (1,1) (from 2).
5. The function is differentiable at point (1,1) (from 3&4)

?
 

1. What is a differentiable vector function?

A differentiable vector function is a mathematical function that maps a set of input values to a set of output values represented as vectors. It is defined as a function that has a well-defined derivative at every point in its domain.

2. How is differentiability of a vector function determined?

The differentiability of a vector function is determined by its ability to have a derivative at every point in its domain. This means that the function must have a unique slope or direction at each point, and the values of the function must be continuous.

3. What is the significance of differentiable vector functions?

Differentiable vector functions are significant in many areas of mathematics, physics, and engineering. They are used to describe and model real-world phenomena, such as motion, forces, and electromagnetic fields. They also play a crucial role in optimization problems, where the goal is to find the maximum or minimum value of a function.

4. How is the derivative of a differentiable vector function calculated?

The derivative of a differentiable vector function can be calculated using the process of differentiation. This involves finding the rate of change of the function at a specific point, which is represented as a vector. The derivative can also be calculated using the gradient, which is a vector that points in the direction of the steepest ascent of the function at a given point.

5. Can every vector function be differentiable?

No, not every vector function is differentiable. A vector function must have a well-defined derivative at every point in its domain to be considered differentiable. Functions that have sharp corners, discontinuities, or undefined points are not differentiable at those points.

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