# Proving a function is not differentiable, when its partials are not continuous

Let $$f(x,y)=\frac{y \sin (3 x)}{\sqrt{x^2+y^2}},$$ and $f(0,0)=0$. I'm trying to prove that it's not differentiable in $(0,0)$. Some my plan was to compute the limit of the definition of differentiability, and check that it doesn't exist. However, when I try to calculate the partial derivatives, I get $$\frac{\partial f}{\partial x}=\frac{3 y \cos (3 x)}{\sqrt{x^2+y^2}}-\frac{x y \sin (3 x)}{\left(x^2+y^2\right)^{3/2}}.$$ Should I just assume that this partial derivative can be extended by continuity, or should I prove it? If I convert to spherical coordinates, I get zero as the limit. I'm not sure if the teacher had this in mind, or I'm doing some mistake...

Any help would be appreciated.

• @GitGud you're right. I've edited. my mistake Jul 10 '16 at 12:48
• OK. So apparently you've found the partials for all points except the origin. The next thing you need to do is find the partials at the origin. Jul 10 '16 at 12:50

## 1 Answer

The function $f$ extends to a continuous function around the origin by setting $f(0,0) = 0$. The partial derivatives at $(0,0)$ are

$$\left.\frac{\partial f}{\partial x}\right|_{(0,0)} = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} 0 = 0,\\ \left.\frac{\partial f}{\partial y}\right|_{(0,0)} = \lim_{h \to 0} \frac{f(0,h) - f(0,0)}{h} = \lim_{h \to 0} 0 = 0$$

so the partial derivatives exist. If $f$ were differentiable at $(0,0)$ then we would have

$$\lim_{(x,y) \to (0,0)} \frac{f(x,y) - f(0,0) - \left.\ \frac{\partial f}{\partial x} \right|_{(0,0)}x - \left.\ \frac{\partial f}{\partial y} \right|_{(0,0)}y}{\sqrt{x^2 + y^2}} = \lim_{(x,y) \to (0,0)} \frac{y\sin(3x)}{x^2+y^2} = 0.$$

However, taking the limit along $x = y$ we have

$$\lim_{x \to 0} \frac{x\sin(3x)}{x^2 + x^2} = \lim_{x \to 0} \frac{\sin(3x)}{2x} = \frac{3}{2}$$

and so $f$ is not differentiable at the origin.

• Please check the improvements I've made and adapt to the rest of your answer, if you wish. Jul 10 '16 at 12:54