# differentiability for trigonometric function in two variable

Given $$f(x,y)=\left\{ \begin{matrix} 0 & (x,y)=(0,0)\\ \frac {\sin\left(x^2-xy \right)}{\vert x \vert} & (x,y) \neq (0,0) \end{matrix} \right.$$

Proved continuity, I have to seek for differentiability. Calculating partial derivative in y by definition gives me a $0/0$ indetermination. Means it is not differentiable or are my calculus wrong?
Thank you

• You should probably use the very definition of partial derivatives instead of rules. – Siminore Jul 14 '16 at 11:10

$$\frac{\partial f}{\partial x} (0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{\sin(h^2)}{h|h|}$$
This limit does not exist; it gives $-1$ when $h$ approaches $0$ from the left and and $1$ when $h$ approaches from the right. This means that the partial derivative wrt $x$ doesn't exist.
Try this: since $${d \over dy} \left({sin(x^2-xy) \over |x|}\right) = -{x \cos(x^2-xy) \over |x|}$$ note that $$-{x \cos(x^2-xy) \over |x|} = \pm{|x|\cos(x^2-xy) \over |x|} = \pm\cos(x^2-xy)$$ and $$\lim_{(x,y) \to 0}\ \pm \cos(x^2-xy) = \pm 1$$ thus the limit does not exist. Since all partial derivative must exist and be continuous at $p$ for a function $f$ to be differentiable at $p$, the function $f$ is not differentiable at $(0,0)$.