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Math Help - Proof that f(x,y) is differentiable and determine the derivate

  1. #1
    Inf
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    Proof that f(x,y) is differentiable and determine the derivate

    Hey guys!

    I am supposed to show that the function f(x,y) = x^2 + y^3 is differentiable and to determine the derivate.

    Do i have to show this with the differential quotient? I know about partial derivates but how to determine the "entire derivate" f'(x,y)?

    I know lots of questions and no solution. Sorry for that, but please give me a clue, some keywords at least.

    Thanks!

    Inf
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    Quote Originally Posted by Inf View Post
    Hey guys!

    I am supposed to show that the function f(x,y) = x^2 + y^3 is differentiable and to determine the derivate.

    Do i have to show this with the differential quotient? I know about partial derivates but how to determine the "entire derivate" f'(x,y)?

    I know lots of questions and no solution. Sorry for that, but please give me a clue, some keywords at least.

    Thanks!

    Inf
    Dear Inf,

    For a multivariable function(a function with two or more variables) the derivative is not defined. Only the partial derivative is defined. So there is no such thing as the "entire derivative."
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  3. #3
    Inf
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    Dear Sudharaka,

    thanks for your quick reply!

    So, these are my partial derivates:
    df(x,y)/dx = 2x
    df(x,y)/dy = 3y^2

    Did i proof now that f is differentiable?

    Inf
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  4. #4
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    Quote Originally Posted by Inf View Post
    Dear Sudharaka,

    thanks for your quick reply!

    So, these are my partial derivates:
    df(x,y)/dx = 2x
    df(x,y)/dy = 3y^2

    Did i proof now that f is differentiable?

    Inf
    If you need to do this by first principles then

    since this is a function from

    f:\mathbb{R}^2 \to \mathbb{R}

    The total derivative will be a 1 by 2 matrix

    The definition is

    \lim_{\mathbf{h} \to \mathbf{0}} \bigg{|}\bigg{|}\frac{f(\mathbf{x}+\mathbf{h})- f(\mathbf{x})- \mathbf{J} \mathbf{h}}{\mathbf{h}} \bigg{|}\bigg{|} =0

    Where

    \mathbf{x} = \begin{bmatrix}x \\ y \end{bmatrix} \quad \mathbf{h} = \begin{bmatrix}h_1 \\ h_2 \end{bmatrix}

    and J is the Jacobian matrix (in this case the gradient of f)

    \mathbf{J} = \begin{bmatrix}2x & 3y^2 \end{bmatrix}

    This will prove that the Jacobian is the derivative.
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  5. #5
    Inf
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    Hey TheEmptySet,

    thank you! After doing some "calculation", i get
    \lim\limits_{h\rightarrow 0}{}\frac{1}{||h||}\cdot(h^3_2 + 3h^2_{2}y + h^2_1) = 0

    ||.|| is a norm, right? which one applies here? euclidean one?

    ||h|| = \sqrt{h^2_1 + h^2_2}

    and if it is this one how to move on?

    Inf
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    Note that

    \bigg|\bigg| \frac{h_1^2+3yh_2^2+h_2^3}{\sqrt{h_1^2+h_2^2}}  \bigg|\bigg| \le \bigg|\bigg|\frac{h_1^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{3yh_2^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{h_2^3}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg|

     \le |h_1|+3|y||h_2| +|h_2^2|
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  7. #7
    Inf
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    Ok, thanks!

    Bunch of questions

    Why are you using a norm on a real number?
    Why are you using inequalities?
    Where is the limit?

    Cheers,
    Inf
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    Behold, the power of SARDINES!
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    Quote Originally Posted by Inf View Post
    Ok, thanks!

    Bunch of questions

    Why are you using a norm on a real number?
    Why are you using inequalities?
    Where is the limit?

    Cheers,
    Inf
    The norm of a scalar is just its absolute value

    The limit itself is hard to evaluate directly but if you use the inequalities above you can use the squeeze theorem

    This gives

    0 \le \bigg|\bigg| \frac{h_1^2+3yh_2^2+h_2^3}{\sqrt{h_1^2+h_2^2}}  \bigg|\bigg| \le \bigg|\bigg|\frac{h_1^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{3yh_2^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{h_2^3}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg|  \le |h_1|+3|y||h_2| +|h_2^2|

    Now if you take the limit as h goes to zero you get

    0 \le \bigg|\bigg| \frac{h_1^2+3yh_2^2+h_2^3}{\sqrt{h_1^2+h_2^2}}  \bigg|\bigg|  \le 0

    So the limit must be zero.
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  9. #9
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    Thank you, now i get it!

    Why
    |h_1|+3|y||h_2| +|h_2^2|
    and not
    |h_1^2|+3|y||h_2^2| +|h_2^3|
    ?

    "h goes to zero" means both h_1 and h_2 go to zero, right?

    Inf
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    Quote Originally Posted by Inf View Post
    Thank you, now i get it!

    Why
    |h_1|+3|y||h_2| +|h_2^2|
    and not
    |h_1^2|+3|y||h_2^2| +|h_2^3|
    ?

    "h goes to zero" means both h_1 and h_2 go to zero, right?

    Inf
    I will do the first one the other two are similar. The problem is that the denominator is also going to zero.

    \bigg|\bigg|\frac{h_1^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg|=|h_1|\bigg|\bigg|\frac{h_1}{\sqrt{h_1  ^2+h_2^2}} \bigg|\bigg|

    Note that

    h_1^2 \le  h_1^2+h_2^2 \implies |h_1| < \sqrt{h_1^2+h_2^2} \iff \frac{|h_1|}{\sqrt{h_1^2+h_2^2}} \le 1

    Now putting this into the above gives

    |h_1|\bigg|\bigg|\frac{h_1}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| \le |h_1|\cdot 1

    Yes you need to let both

    h_1 \to 0 \text{ and } h_2 \to 0
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  11. #11
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    Thank you, now i get it!

    Why
    |h_1|+3|y||h_2| +|h_2^2|
    and not
    |h_1^2|+3|y||h_2^2| +|h_2^3|
    ?

    "h goes to zero" means both h_1 and h_2 go to zero, right?

    Inf
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    Quote Originally Posted by Sudharaka View Post
    Dear Inf,

    For a multivariable function(a function with two or more variables) the derivative is not defined. Only the partial derivative is defined. So there is no such thing as the "entire derivative."
    That's simply not true- but of course, you have to generalize the notion of "derivative". If f is a function from R^n to R^m then the derivative of f, at point p, is the linear function from R^n to R^m, that best approximates f in a neighborhood of p. Specifically, we say that such a function is differentiable at p if and only if there exist a linear function L and a function \epsilon(x), from R^n to R^m, such that
    f(x)= f(p)+ L(x- p)+ \epsilon(x)
    \lim_{x\to p}\frac{\epsilon(x)}{||x- p||}= 0

    Notice the the derivative is defined as a linear transformation. In the case of R^1 to R^1, it would be the function y= \frac{df}{dx}(p)x where \frac{df}{dx}(p) is the "usual" derivative at p. In the case of R^2 to R^1 the linear transformation is maps (p_x, p_y) to \frac{\partial f}{\partial x}p_x+ \frac{\partial f}{\partial y}p_y. In other words it is the dot product of \nabla f and p. That's why most texts treat \nabla f as being "the" derivative.
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  13. #13
    Inf
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    Ok, now i finally understand! thank you guys!
    Sorry for the double post above..
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