## A function non-differentiable at (0,0) but all traces through (0,0) are differentiable

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Consider the function $f(x,y) = \begin{cases} -\dfrac{xy^2}{x^2+y^2} & x \neq 0 \\ 0 & x = 0\end{cases}$ Traces along all directions through $$(0,0)$$ have tangent lines (i.e. are differentiable) at $$(0,0)$$. However, there is no single plane that contains all these tangent lines. To see this, we can look at a few of these traces.

A trace passing through $$(0,0)$$ along the direction $$\theta$$ (measured from the $$x$$-axis) can be represented by $$(x,y) = (t\cos\theta,t\sin\theta)$$. By plugging into $$f$$ you get $f(t\cos\theta,t\sin\theta) = -t\cos\theta\sin^2\theta$ after simplification. It follow that $$df/dt = -\cos\theta\sin^2\theta$$.

Along the $$x$$-axis ($$y = 0, \theta = 0$$), $\left.\dfrac{df}{dt}\right|_{t=0} = \left.\dfrac{\partial f}{\partial x}\right|_{(0,0)} = 0.$ Along the $$y$$-axis, ($$x = 0, \theta = \pi /2$$), $\left.\dfrac{df}{dt}\right|_{t=0} = \left.\dfrac{\partial f}{\partial y}\right|_{(0,0)} = 0.$ Based on these partial derivatives, we may conclude that $$z = 0$$ is the tangent plane. However, if we follow the trace along $$\theta = \pi/4$$, then we find that the slope of this tangent line is $\left.\dfrac{df}{dt}\right|_{t=0} = -\dfrac{1}{\sqrt{8}} \neq 0$ at $$(0,0)$$ and this tangent line certainly does not lie on the plane $$z = 0$$.

The following diagram shows the tangent lines through $$(0,0)$$ along various directions.

Since there is no plane that can contain all these tangent lines, we conclude that a tangent plane at $$(0,0)$$ does not exist and $$f$$ is not differentiable at $$(0,0)$$ even derivative exists in all directions.

If a function is non-differentiable at a point, then some of the partial derivatives must be discontinuous at that point. This is illustrated in the following diagram. For the above $$f$$, $\dfrac{\partial f}{\partial x} = \begin{cases} \dfrac{y^2(x^2-y^2)}{(x^2+y^2)^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0)\end{cases}$ Along $$x = 0$$, $\left.\dfrac{\partial f}{\partial x}\right|_{(0,y)} = \begin{cases} -1 & y \neq 0 \\ 0 & y = 0 \end{cases}$ which says $$\partial f/\partial x$$ is discontinuous at $$(0,0)$$.

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