# Conflicting result in derivative of composite function

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1. Oct 10, 2016

### toforfiltum

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)$. I don't know, I'm just guessing here. Can someone help me? Thanks. 2. Oct 10, 2016 ### Ray Vickson Don't guess; verify. You now know the definition of differentiability at$(x,y) = (0,0)$, and can check whether or not your$f$verifies that definition. If you cannot check it in cartesian coordinates, try switching to polar coordinates. 3. Oct 11, 2016 ### toforfiltum Ah, switching to polar coordinates tells me that$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)$? 4. Oct 11, 2016 ### toforfiltum @Ray Vickson Yes, what I say may be true. At$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$is not differentiable at 0. Hence chain rule cannot be applied here. Is this right? 5. Oct 11, 2016 ### toforfiltum I realised I have mixed up between continuity and differentiability. In post #3, I should have said continuous, not differentiable.$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)$. This explains why the chain rule does not work, but it still baffles me why the differentiation of the composite function works. 6. Oct 11, 2016 ### Ray Vickson Why would you say that the reason$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$verifies that definition at the origin. 7. Oct 11, 2016 ### toforfiltum That's not what I said, my TA said it. That didn't make sense to me, because like you said, it is defined at that point. And I did use the definition of differentiability in the post above to show that it is not differentiable at that point. Perhaps I will type everything out this time. Since partial derivatives at$(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)$. 8. Oct 11, 2016 ### Ray Vickson OK, that is good enough. Another way is to ask: are there constants$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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