# Directional derivatives — implied continuity or not?

Consider the function : $$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto \begin{cases} 0 & \text{for } (x,y)=(0,0) \\ \frac{x^2y^2}{x^2y^2+(y-x)^3} & \text{for } (x,y) \neq (0,0) \end{cases}$$ Is it differentiable and continuous at $$0$$?

I calculated the directional derivative which is equal to $$0$$ for all $$u$$ not equal to $$0$$. If all directional derivatives equal to $$0$$, does this imply differentiability and continuity? I cannot seem to find an answer online. But, if not, then how would i find if it is differentiable or not?

The answer is no. Even if a function has a directional derivative for any direction, the possibility that the function is not continuous is still opened. The idea is that the directional derivative only captures the behavior of a function at a point along a line, so it could fail to catch its continuity or differentiability along other curves.

For example, consider $$f(x,y)=\begin{cases} \frac{x^2y}{x^4+y^2} & \text{if }(x,y)\neq(0,0), \\ 0 & \text{if }(x,y)=(0,0).\end{cases}$$ and take $$(x,y)=(t,t^2)$$. Then $$f(t,t^2)=\frac{1}{2}$$. However, $$f$$ has a directional derivative at the origin for any direction.

In the case of your function, take $$(x,y)=(t,t^2+t)$$. Then $$f(t,t^2+t) = \frac{t^4(t+1)^2}{t^4(t+1)^2+t^6}=\frac{1}{1+t^2(1+t)^{-1}}.$$ It tends to $$1$$ if $$t\to 0$$. However, $$f(0,0)=0$$, so $$f$$ is not even continuous at the origin.

• I think, Even the directional derivative does not exist with respect to the vector $u=(h,h)$ Commented Jan 26, 2020 at 9:11
• @Gune You are right in the case of the given function. I need to replace the example. Commented Jan 26, 2020 at 9:16

Claim: The directional derivative does not exist with respect to the vector $$u=(h,h)$$ for $$h\neq0$$.

Proof:

$$\begin{split}f'(0,u)&=\lim\limits_{t\to0}\frac{f(0+tu)-f(0)}{t}\\ &=\lim\limits_{t\to0}\frac{\frac{t^4h^4}{t^4h^4+t^3(h-h)^3}-0}{t}\\ &=\lim\limits_{t\to0}\frac{1}{t}\\ \end{split}$$

This limit does not exist. Hense $$f$$ is not differentiable at(0,0).