Question

Consider f:R² to R defined by f(x,y) =(x+y)/(√2) if x=y and f(x,y) =0 otherwise ,show.that f_x(0,0) =f_y(0,0)=0 and D_uf(0,0)=1 ,where ,u=(1/√2,1/√2) Deduce that f is not differentiable at (0,0)

Answer 1

The directional derivatives are $f_x(0,0)=0'=0$ and $f_y(0,0)=0'=0$. The derivative in direction $u=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ is $D_u(0,0)=\frac{1}{\sqrt{2}}f_x+\frac{1}{\sqrt{2}}f_y=\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}=1$. The derivatives in directions $x$, $y$ and $u$ do not coincide. Therefore, the function is not differentiable at (0,0).