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

  • $\begingroup$ You should probably use the very definition of partial derivatives instead of rules. $\endgroup$ – 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.

The nonexistence of the partial derivatives at a point implies non-differentiability of the function at the point.


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 $(0,0)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.