Let $f: \mathbb R^2 \rightarrow \mathbb R $ be defined as $$ f(x,y) = \begin{cases} \dfrac {xy^2}{x^2+y^4} & x\ne 0 \\\\ 0 & x=0 ~ \end{cases} $$

Let $D_u f(0,0)$ denote the directional derivative of $f$ at $(0,0)$ in the direction $u = (u_1,u_2) \ne (0,0).$ then $f$ is :

$(i) $ continuous at $(0,0)$ and $D_uf(0,0)$ exists for all $u$.

$(ii) $ continuous at $(0,0)$ but $D_uf(0,0)$ does not exist for some $u \ne (0,0)$

$(iii) $ not continuous at $(0,0)$ and $D_uf(0,0)$ exists for all $u$.

$(iv) $ not continuous at $(0,0)$ and $D_uf(0,0)$ does not exist for some $u \ne (0,0)$


This function is not continuous at $(0,0)$ as along the curve $x = my^2 : f(x,y) = \dfrac {m}{1+m^2}$ which depends on the value of $m$.

In some direction $\theta : x =a \cos \theta, y = a \sin \theta$ and hence the directional derivative at $(0,0)$ in the direction of $\theta$ along a curve of length $a \rightarrow 0 $ will be given as :

$D_u(0,0) =\lim_{a \rightarrow 0} \dfrac {a \cos \theta \cdot a^2 \sin^2 \theta} {a \big (a^2 \cos^2 \theta + a^4 \sin^4 \theta \big )} = \lim_{a \rightarrow 0} \dfrac {\cos \theta \sin^2 \theta}{\cos^2 \theta + a^2 \sin^4 \theta} = \dfrac{\sin^2 \theta}{\cos \theta}$

which does not exist for $\theta = \dfrac {\pi} {2}$

Hence, the function is not continuous at $(0,0)$ and for $u=(0,1), D_u(0,0)$ does not exist either.

Hence, the correct option should be $(iv)$.

Could someone tell me if I am correct?

Thank you!

  • $\begingroup$ $f: \mathbb R^2 \rightarrow \mathbb R$ $\endgroup$ – Bumblebee Jun 16 '15 at 9:44

Your argument that $f$ is not continuous at $(0,0)$ is fine.

Your computation of $D_\theta f(0,0)$ is not applicable when $\theta=\pm{\pi\over2}$. You have to treat these cases separately and will obtain $D_{\pm\pi/2}f(0,0)=0$.

  • $\begingroup$ When $\theta= \dfrac {\pi}{2}$ from origin, then $x=0.$ substituting in the given function, we get $f(x,y)= 0.$ hence directional derivative along y axis is $0$. That means directional derivative exists everywhere? $\endgroup$ – MathMan Jun 16 '15 at 10:01

Hint: Note that the definition of the directional derivative in the direction $\mathbf{u} = (u_1,u_2)$ is given by $$D_\mathbf{u}f(x,y) \equiv \lim_{h \to 0} \frac{f(x+hu_1,y+hu_2)-f(x,y)}{h}. $$ Try to work out this expression using the function at hand.


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.