0
$\begingroup$

Let a function be given by
$f(0,0) = 0$ and
$f(x,y) = \frac {sin(x^3+y^3)}{x^2+y^2}$ for $(x,y)\neq (0,0)$.
I have already proved that all directional derivatives exist.
Is the function differentiable at $(0,0)$?

$\endgroup$
1
$\begingroup$

No. If it is differentiable then the derivative is necessarily given by $A(x,y)=x+y$. (By looking at the partial derivatives). To get a contradiction from this consider $\frac {f(x,y)-A(x,y)} {\sqrt {x^{2}+y^{2}}}=\frac {\sin (x^{3}+y^{3}) -(x+y)} {\sqrt {x^{2}+y^{2}}}$. I will let you show that $\frac {\sin (x^{3}+y^{3})} {\sqrt {x^{2}+y^{2}}} \to 0$ and $\frac {x+y} {\sqrt {x^{2}+y^{2}}} $does not tend to $0$.

$\endgroup$

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.