# Linear dependence of directional derivative problem

Real analysis problem on directional derivatives.

For $$(x,y)\neq(0,0)$$, let $$f(x,y)=\frac{yx^6+y^3+x^3y}{x^6+y^2}$$ and let $$f(x,y)=0$$ when $$(x,y)=(0,0)$$

a) Show that all directional derivatives of f exist at $$(0,0)$$, and depend linearly on the vector we differentiate along.

b) Show that nevertheless, f is not differentiable at $$(0,0)$$.

What I have so far:

a) Set $$u=(u_1,u_2)$$ such that $$|u|=1$$. By definition $$D_u f=\lim_{h\to 0}\frac{f((x,y)+h(u_1,u_2))-f(x,y)}{h}$$

So then we see $$D_u f(0,0)=\lim_{h\to 0}\frac{f((0,0)+h(u_1,u_2))-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{hu_2(hu_1)^6+(hu_2)^3+(hu_1)^3hu_2}{(hu_1)^6+(hu_2)^2}}{h}=\lim_{h\to 0}\frac{h^7u_2 u_1^6+h^3u_2^3+h^4u_1^3u_2}{h^7u_1^6+h^3u_2^2}=\lim_{h\to 0}\frac{h^4u_2 u_1^6+u_2^3+hu_1^3u_2}{h^4u_1^6+u_2^2}=\frac{u_2^3}{u_2^2}=u_2$$

Hence all directional derivatives of f exist at (0,0). But how would I show that they depend linearly on the vector we differentiate along?

b) First set $$y=0$$, then $$\lim_{(x,y)\to(0,0)} f(x,y)=0$$. Next set $$y=x^3$$, then $$\lim_{(x,y)\to(0,0)} f(x,y)=\frac{1}{2}$$. Since the limits don't match, the function is not continuous at the origin and thus not differentiable at the origin.

• The function $(u_1,u_2) \to u_2$ is linear. – zhw. Mar 4 at 22:12