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Condition for differentiability

  1. Jun 27, 2010 #1
    What's the condition for f(x,y) to be differentiable in its domain?

    I googled for it but couln't find...

    Thanks in advance.
  2. jcsd
  3. Jun 27, 2010 #2
    Are you asking for partial or total differentiability?
    Do you know the definition of differentiability for a function f(x) of one argument?
  4. Jun 27, 2010 #3
    What you mean by "total differentiability"? I thought that when derivating a function with multiple variables I should diferentiate it for x and then for y.

    For a function of one argument to be differentiable it needs to be countinous in its domain?
  5. Jun 27, 2010 #4

    No. A function [itex]x\mapsto f(x)[/itex] is differentiabe at [itex]x_0[/itex] if the limit
    \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}
    exists. It is true that if a function is not continuous at a point is certainly not differentiable there.

    For functions with more than one argument the partial derivatives are the usual derivatives with all but one variable held constant, while the total derivative is a linear operator which best approximates the function as a whole.

    There usually is no general criterion you can check in order to see if a function is differentiable. Of course, for a large class of functions we know that they are differentiable and we also know that differentiability is preserved under algebraic operation and compositions (unless you dived by zero...) so seeing if a given function is differentiable is often easy.

    What is the context of your question?
  6. Jun 27, 2010 #5
    Uhmmm the question is:
    "Verify if each function is differentiable at the points of its domain."
    a)f(x,y) = 2x - y
    b)f(x,y) = ln(2x²+3y²)
  7. Jun 28, 2010 #6


    Staff: Mentor

    Do you know what the domain is for each of these functions?
  8. Jun 28, 2010 #7
    Certainly, if you are asked to prove differentiability of these two functions you are given some sort of definition of this concept.

    I suggest you review this definition and then we can see where you might have problems verifying it for the two examples you gave.
  9. Jun 28, 2010 #8


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    Moderator's note: thread moved to Homework forum from Calculus & Analysis
  10. Jun 28, 2010 #9
    For the first one, the domain is R², right? And for the second, as there is no log of numbers equal or below zero, we have that x²+y²>0. Isn't that right?

    The teacher told us to use a certain book to study, but this is not there...

    Thanks again.
  11. Jun 28, 2010 #10


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    It's hard to believe that the definition of differentiability of a function of two variables is not given in any textbook on Calculus of several variables. The existance of partial derivatives of a function at a given point is NOT enough to prove that the function is "differentiable" there. The standard definition is this:

    F(x, y), a function from [itex]R^2[/itex] to R (you can extend to as many variables as you like) is "differentiable at [itex](x_0, y_0)[/itex] if and only if there exist a linear function L(x,y) from [itex]R^2[/itex] to R and a function [itex]\epsilon(x,y)[/itex] from [[itex]R^2[/itex] to R such that
    [tex]F(x,y)= F(x_0, y_0)+ L(x-x_0, y-y_0)+ \epsilon(x-x_0, y-y_0)[/tex]
    [tex]\lim_{(x,y)\to (x_0,y_0)} \frac{\epsilon(x-x_0, y-y_0)}{\sqrt{(x-x_0)^2+ (y-y_0)^2}}= 0[/tex]

    Roughly, that says that L(x,y) (the derivative of F(x,y) at [itex](x_0, y_0)[/itex] is the "best" linear approximation to F(x,y) in the neighborhood of [itex](x_0, y_0)[/itex]. If you think of z= F(x,y) as defining a surface in [itex]R^3[/itex], that says that the surface has a tangent plane at the point.

    That can be hard to use. Probably what you want to use is a theorem normally proved immediatly after that definition is introduced:
    "F(x,y) is differentiable at [itex](x_0, y_0)[/itex] if and only if the partial derivatives [itex]\partial F/\partial x[/itex] and [itex]\partial F/\partial y[/itex] exist and are continuous in some neighbohood of [itex](x_0, y_0)[/itex]."
  12. Jun 28, 2010 #11


    Staff: Mentor

    Yes, as far as it goes. For the second function, how else could you describe the domain?
  13. Jun 28, 2010 #12
    But how can I find this [itex](x_0, y_0)[/itex] when the exercise asks if its differentiable in R²?

    For the second function, I can also say that the domain is all the circunferences with a radius bigger than 0... does it helps in anything?

  14. Jun 28, 2010 #13


    Staff: Mentor

    A function is differentiable in R2 if it is differentiable at each point in R2.
    I know what you're trying to say, but you're not saying it very well. For this problem, it's easier to say where it is not defined.
  15. Jun 28, 2010 #14
    Perhaps I figured that out for the second one:
    I should find the partial derivative of the function using limits when x and y tend to 0?

    Yet the first one isn't very clear to me...
  16. Jun 28, 2010 #15


    Staff: Mentor

    Partial derivatives. There are two of them.

    The function in the 2nd problem is not defined at (0, 0), so its partial derivatives are not going to be defined there, either.
    But that's the easier one!
    You have f(x, y) = 2x - y. What are the two partials? Where are they continuous?
  17. Jun 28, 2010 #16
    Sorry, I misspelled the word.

    df/dx = 2 and df/dy = -1, and they are continuous in R²... right?
  18. Jun 28, 2010 #17


    Staff: Mentor

  19. Jun 28, 2010 #18
    Now I got it! Thanks a lot everyone for your help!
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