Differentiation Aggravation

Calculus Level 5

$f(x,y) = \begin{cases} \dfrac{x^2y^2}{x^2+y^2} & \text{ if } (x,y)\neq(0,0) \\ \; \; \; \; \; 0 & \text{ if } (x,y)=(0,0) \end{cases} \\ g(x,y) = \begin{cases} \dfrac{xy}{x^2+y^2} & \text{ if } (x,y)\neq(0,0) \\ \; \; \; \; \; 0 & \text{ if } (x,y)=(0,0) \end{cases}$

Consider the functions $f,g : \mathbb{R}^2 \to \mathbb{R}$ as described above.

Are these functions differentiable at $(x,y) = (0,0)$ ?

Only

f

None of them are differentiable at

(x,y)=(0,0)

Both

f

and

g

Only

g

2 solutions

Guilherme Dela Corte
Dec 29, 2016

If $h(x,y)$ is differentiable at $(x_0, y_0)$ , then both $\frac{\partial h}{\partial x} (x_0, y_0)$ and $\frac{\partial h}{\partial y} (x_0, y_0)$ exist, such that

$h(x_0 +\Delta x, y_0 + \Delta y ) = h(x_0, y_0) + \frac{\partial h}{\partial x} (x_0, y_0) \cdot \Delta x + \frac{\partial h}{\partial y} (x_0, y_0) \cdot \Delta y + R_h(\Delta x , \Delta y )$ $\lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_h(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} = 0.$

We can show that $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_f(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} = 0$ , but $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_g(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|}$ is indeterminate, and thus $g$ is not differentiable at $(0,0)$ .

Can you add the details of the final "We can show that"? It feels like you're building up the theory to help someone understand, but not completing the final step for them to see why we reach that conclusion.

Calvin Lin Staff - 4 years, 5 months ago

It is easy to calculate and see that $\frac{\partial f}{\partial x} (0,0) = \frac{\partial f}{\partial y} (0,0) = \frac{\partial g}{\partial x} (0,0) = \frac{\partial g}{\partial y} (0,0) = 0$ .

Thus, we see that $R_f (\Delta x, \Delta y) = \dfrac{(\Delta x)^2(\Delta y)^2}{(\Delta x)^2+(\Delta y)^2}$ and $R_g (\Delta x, \Delta y) = \dfrac{\Delta x \Delta y}{(\Delta x)^2+(\Delta y)^2}$ .

First, let's prove that $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_f(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} = 0$ , where $\dfrac{R_f(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} = \dfrac{(\Delta x)^2(\Delta y)^2}{((\Delta x)^2+(\Delta y)^2)^{\frac{3}{2}}}$ .

$0 \leq \dfrac{(\Delta x)^2(\Delta y)^2}{((\Delta x)^2+(\Delta y)^2)^{\frac{3}{2}}}\leq \dfrac{(\Delta x)^2(\Delta y)^2}{((\Delta x)^2+(\Delta y)^2)^{\frac{3}{2}}} + {\color{#456461} \frac{1}{2} \cdot \dfrac{(\Delta x)^4+(\Delta y)^4}{((\Delta x)^2+(\Delta y)^2)^{\frac{3}{2}}}} = \frac{\sqrt{(\Delta x)^2+(\Delta y)^2}}{2} \Rightarrow$

$\Rightarrow \lim_{(\Delta x, \Delta y) \to (0,0)} 0 \leq \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_f(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} \leq \lim_{(\Delta x, \Delta y) \to (0,0)} \sqrt{(\Delta x)^2+(\Delta y)^2}$

Through Squeeze Theorem , $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_f(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} = 0$ . Because of this, $f$ is differentiable at $(0,0)$ .

Now we will show that $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_g(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} = \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{\Delta x\Delta y}{((\Delta x)^2+(\Delta y)^2)^{\frac{3}{2}}} \neq 0$ .

Let $\Delta x = \Delta y$ . From this approach, we see that $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{\Delta x\Delta y}{((\Delta x)^2+(\Delta y)^2)^{\frac{3}{2}}} = \lim_{\Delta x \to 0} \dfrac{(\Delta x)^2}{2\sqrt{2} \cdot \left | \Delta x \right |^3} = \lim_{\Delta x \to 0} \dfrac{1}{2\sqrt{2} \cdot |\Delta x|} .$

However, $\displaystyle \lim_{\Delta x \to 0} \dfrac{1}{2\sqrt{2} \cdot |\Delta x|} = +\infty$ , not $0$ .

Thus we have shown that $\displaystyle \lim_{(\Delta x, \Delta y) \to (0,0)} \dfrac{R_g(\Delta x, \Delta y)}{\left \| (\Delta x, \Delta y) \right \|} \neq 0$ . Because of this, $g$ is not differentiable at $(0,0)$ .

Guilherme Dela Corte - 4 years, 5 months ago

Hjalmar Orellana Soto
Dec 29, 2016

A function $h(x,y)$ is differentiable at $(x_{0},y_{0})$ iif: $\lim_{(x,y)\to(x_{0},y_{0})}\frac{h(x,y)-h(x_{0},y_{0})-\frac{\partial h}{\partial x} (x_{0},y_{0}) \cdot x - \frac{\partial h}{\partial y}(x_{0},y_{0}) \cdot y}{||(x+x_{0},y+y_{0})||}=0$ Then, we find the following values: $\frac{\partial f}{\partial x} (0,0)=\lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0} \frac{0}{h}=0=\frac{\partial f}{\partial y}(0,0)=\frac{\partial g}{\partial x}(0,0) =\frac{\partial g}{\partial y}(0,0)$ So, replacing in the definition above, we have: $\lim_{(x,y)\to (0,0)}\frac{x^2 y^2}{(x^2+y^2)^\frac{3}{2}}=\lim_{r\to 0 \wedge \theta free} \frac{r^4\cdot \frac{1}{2} \sin(2\theta)}{r^3}=0$ $\Rightarrow f(x,y)$ is differentiable at $(0,0)$ $\lim_{(x,y)\to (0,0)}\frac{xy}{(x^2+y^2)^\frac{3}{2}}=\lim_{r\to 0 \wedge \theta free} \frac{r^2 \cdot \frac{1}{2} \sin(2\theta)}{r^3}$ Which depends on the value of $\theta$ , so doesn't exist. $\Rightarrow g(x,y)$ is not differentiable at $(0,0)$

Can you explain what does the limit $r \rightarrow 0\wedge \theta free$ means?

Calvin Lin Staff - 4 years, 5 months ago

Sorry, is a polar change of variables, with $r \to 0$ and with a free $\theta$ , this is, $\theta$ can take any real value

Hjalmar Orellana Soto - 4 years, 5 months ago

Differentiation Aggravation

2 solutions

0 pending reports