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Conflicting result in derivative of composite function

  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Let $$f(x,y)=\begin{cases} \frac{x^2y}{x^2+y^2} \space & \text{if} \space(x,y)\neq(0,0)\\0 \space & \text{if} \space(x,y)=(0,0)\end{cases}$$

    a) Use the definition of the partial derivative to find ##f_x(0,0)## and ##f_y(0,0)##.

    b) Let a be a nonzero constant and let ##x(t)=(t,at)##. Show that ##f\circ x## is differentiable, and find ##D(f\circ x)(0)## directly.

    c) Calculate ##Df(0,0)Dx(0). How can you reconcile your answer in part b) and the chain rule?

    3. The attempt at a solution
    a) By going to the limit definition ##f_x(0,0)=0## and ##f_y(0,0)=0##.

    b) $$f \circ x(t,at)=\frac{t^2(at)}{t^2(1+a^2)}$$
    $$=\frac {at}{1+a^2}$$

    Hence,$$D(f \circ x)=\frac{a}{1+a^2}$$ for all values of t.

    c) According to answer in (a), $$Df(0,0)=(0,0)$$
    $$Dx(0)= \left(\frac 1 a\right)$$
    The product of both matrices yields zero.

    I think that this conflicting result shows that at ##(0,0)##, there is no differentiability. Maybe along the line ##x(t)##, there is a derivative, but not for other paths going to ##(0,0)##.

    I don't know, I'm just guessing here. Can someone help me?

    Thanks.
     
  2. jcsd
  3. Oct 10, 2016 #2

    Ray Vickson

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    Don't guess; verify.

    You now know the definition of differentiability at ##(x,y) = (0,0)##, and can check whether or not your ##f## verifies that definition. If you cannot check it in cartesian coordinates, try switching to polar coordinates.
     
  4. Oct 11, 2016 #3
    Ah, switching to polar coordinates tells me that ##f## is indeed differentiable at ##(0,0)##, because the limit exists and is equal to zero. This invalidates my argument above.

    Does this mean that ##x## may not be differentiable at ##(0,0)##?
     
  5. Oct 11, 2016 #4
    @Ray Vickson Yes, what I say may be true. At ##x(0)##, the limits of the 2 functions are not equal to each other.

    For ##t##, $$\lim_{h \to 0}\frac{x(h)-x(0)}{h}$$
    $$=\frac{h-0}{h}=1$$

    For ##at##, $$\lim_{h \to 0}\frac{x(h)-x(0)}{h}$$
    $$=\frac{ah-0}{h}=a$$

    Since the limits are not equal to each other, ##x## is not differentiable at 0.

    Hence chain rule cannot be applied here.

    Is this right?
     
  6. Oct 11, 2016 #5
    I realised I have mixed up between continuity and differentiability. In post #3, I should have said continuous, not differentiable.

    ##D(f \circ x)(0)## works because the input of ##f(x,y)## now becomes ##f(t,at)##, and ##x(t)## is defined for all values of ##t##. So maybe this composition helps to sidestep the condition where ##f(x,y)=0## if ##(x,y)=(0,0)##.

    According to my TA,the chain rule doesn't work here because ##f## is not differentiable at ##(0,0)## since it is undefined at ##(0,0)##.

    And I think I can prove this.

    $$\lim_{(x,y) \to (0,0)} \frac{f(x,y) -0}{\|(x,y)-(0,0)\|}$$
    $$=\frac{x^2y}{(x^2+y^2)^\frac{3}{2}}$$
    When function is approached along ##y=0## and ##x=0##, limit is zero.

    But when function is approached along ##y=mx## as ##x \to 0##, a non zero value is obtained. Hence function is not differentiable at ##(0,0)##.

    This explains why the chain rule does not work, but it still baffles me why the differentiation of the composite function works.
     
  7. Oct 11, 2016 #6

    Ray Vickson

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    Why would you say that the reason ##f## is not differentiable at ##(0,0)## is that it is undefined at ##(0,0)##. That is false reasoning: ##f## is defined at the origin---just go back and look at the definition of ##f## you gave in Post #1. It is true that ##f## is not differentiable at the origin, but for a different reason. Can you tell us why? Look at the definition of differentiability, and see if your ##f## verifies that definition at the origin.
     
  8. Oct 11, 2016 #7
    That's not what I said, my TA said it. That didn't make sense to me, because like you said, it is defined at that point.

    And I did use the definition of differentiability in the post above to show that it is not differentiable at that point. Perhaps I will type everything out this time.

    Since partial derivatives at ##(0,0)##, are both zero and function is defined as zero at ##(0,0)##,

    $$\lim_{(x,y) \to (0,0)} \frac{f(x,y)-0}{\|(x,y)-(0,0)\|}$$
    $$=\frac{x^2y}{(x^2+y^2)^\frac{3}{2}}$$

    Along the line ##y=0##, ##\lim_{x \to 0}=0##

    Along the line ##x=0##, ##\lim_{y \to 0}=0##

    Along the line ##y=mx## as ##x \to 0##, $$\lim_{y=mx \to 0} \frac{mx^3}{x^3(1+m^2)^\frac{3}{2}} \neq 0$$

    Hence, function is not differentiable at ##(0,0)##.
     
  9. Oct 11, 2016 #8

    Ray Vickson

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    OK, that is good enough.

    Another way is to ask: are there constants ##a## and ##b## such that ##f(x,y) = f(0,0) + a x + by + e(x,y)##, where the "error" term ##e(x,y)## is of higher order in small ##\|(x,y)\|## (that is, where ##e(x,y)/ \|(x,y)\| \to 0## as ##(x,y) \to (0,0)##)? If you had such constants ##a,b##, the would, necessarily, have to equal ##f_x(0,0)## and ##f_y(0,0)##, respectively, because ##f(x,0) - f(0,0) = ax + e(x,0)##, so ##[f(x,0)-f(0,0)]/x \to a ## as ##x \to 0##, etc. However, we have seen already that ##f_x = f_y = 0## at the origin, so if ##f## were differentiable at the origin we would have to have ##f(x,y) = 0 x + 0 y + e(x,y) = e(x,y)## of "higher order"; that is, we would have to have ##f(x,y)/\|(x,y)\| \to 0 ## as ##\|(x,y)\| \to 0##. Is that true? Well, using polar coordinates we have ##f(x,y) = r \cos^2(\theta) \sin(\theta)## and, of course, ##\|(x,y)\| = r##, so ##f(x,y)/\|(x,y)\| = \cos^2(\theta) \sin(\theta)##. For most ##\theta## this does not go to 0 as ##r \to 0##.
     
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