Very Easy Differential Question

In summary, we can conclude that the first partial derivatives w.r.t x and w.r.t y exist for the given function f(x,y) = (xy)/(x^2 + y^2).
  • #1
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Homework Statement


We have [tex]f(x,y) = \frac{xy}{x^2+y^2}[/tex]
Show that the first partial derivative w.r.t. x and w.r.t y exist


Homework Equations



f(x+dx,y)-f(x,y) = a(dx) + o(dx) where a is some number and o(dx)(not o multiplied by dx rather a 'function', if so to say, o of dx) is such that o(dx)/dx goes to 0 as dx goes to 0



The Attempt at a Solution



I am stuck at the very beginning by just substituting for f(x+dx,y)-f(x,y) and then I don't know what to do. I am not able to tell which part is a(dx) and which part is o(dx). This is absurdly easy and I've done such things before but I am having a block and for some reason can't get past the algebra. Thanks
 
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  • #2
for any help.

it is important to have a clear understanding of the problem at hand before attempting to solve it. In this case, the first step would be to define the function f(x,y) and its domain. This can be done by stating that f(x,y) is a real-valued function defined on the set of all real numbers except for (0,0), since the denominator in the function is undefined at that point.

Next, we can rewrite the given function as f(x,y) = xy(x^2 + y^2)^-1. This form makes it easier to see that the function is continuous and differentiable everywhere except for (0,0), since the denominator is never equal to 0 for any other point in the domain.

To show that the first partial derivatives exist, we can use the definition of partial derivatives:

f_x(x,y) = lim(h->0) [f(x+h,y) - f(x,y)]/h and f_y(x,y) = lim(h->0) [f(x,y+h) - f(x,y)]/h

Substituting in the given function, we get:

f_x(x,y) = lim(h->0) [xy(x+h)^2 + y^2 - xy(x^2 + y^2)]/[h(x^2 + y^2)(x+h)^2] = lim(h->0) [2xyh + yh^2]/[h(x^2 + y^2)(x+h)^2]

= lim(h->0) [2xy + yh]/[(x^2 + y^2)(x+h)^2] = [2xy]/[(x^2 + y^2)^2] = [2xy]/[(x^2 + y^2)^2]

Similarly, f_y(x,y) = [2xy]/[(x^2 + y^2)^2]

Both of these partial derivatives exist and are continuous everywhere except for (0,0), since the limit exists and is finite for all points in the domain.

In conclusion, we have shown that the first partial derivatives of f(x,y) exist for all real numbers except for (0,0). This means that the function is differentiable everywhere except for (0,0).
 

What is a differential equation?

A differential equation is an equation that relates a function to its derivatives. It describes the relationship between the rate of change of a variable and the value of the variable itself.

What is the easiest type of differential equation?

The easiest type of differential equation is a separable differential equation. This type of equation can be solved by separating the variables on each side of the equation and integrating both sides.

What is the order of a differential equation?

The order of a differential equation is the highest derivative present in the equation. For example, a first-order differential equation contains only first derivatives, while a second-order differential equation contains second derivatives.

How do I solve a very easy differential equation?

To solve a very easy differential equation, you can use the method of separation of variables or the method of integrating factors. These methods involve manipulating the equation to separate the variables and then integrating both sides to find the solution.

What are some real-life applications of differential equations?

Differential equations are used in many fields of science, such as physics, chemistry, biology, and engineering. They are used to model and understand various natural phenomena, including population growth, chemical reactions, and electrical circuits.

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