3.II.9C

Vector Calculus
Part IA, 2004

For a function f:R2Rf: \mathbb{R}^{2} \rightarrow \mathbb{R} state if the following implications are true or false. (No justification is required.)

(i) ff is differentiable f\Rightarrow f is continuous.

(ii) fx\frac{\partial f}{\partial x} and fy\frac{\partial f}{\partial y} exist f\Rightarrow f is continuous.

(iii) directional derivatives fn\frac{\partial f}{\partial \mathbf{n}} exist for all unit vectors nR2f\mathbf{n} \in \mathbb{R}^{2} \Rightarrow f is differentiable.

(iv) ff is differentiable fx\Rightarrow \frac{\partial f}{\partial x} and fy\frac{\partial f}{\partial y} are continuous.

(v) all second order partial derivatives of ff exist 2fxy=2fyx\Rightarrow \frac{\partial^{2} f}{\partial x \partial y}=\frac{\partial^{2} f}{\partial y \partial x}.

Now let f:R2Rf: \mathbb{R}^{2} \rightarrow \mathbb{R} be defined by

f(x,y)={xy(x2y2)(x2+y2) if (x,y)(0,0)0 if (x,y)=(0,0)f(x, y)= \begin{cases}\frac{x y\left(x^{2}-y^{2}\right)}{\left(x^{2}+y^{2}\right)} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{cases}

Show that ff is continuous at (0,0)(0,0) and find the partial derivatives fx(0,y)\frac{\partial f}{\partial x}(0, y) and fy(x,0)\frac{\partial f}{\partial y}(x, 0). Then show that ff is differentiable at (0,0)(0,0) and find its derivative. Investigate whether the second order partial derivatives 2fxy(0,0)\frac{\partial^{2} f}{\partial x \partial y}(0,0) and 2fyx(0,0)\frac{\partial^{2} f}{\partial y \partial x}(0,0) are the same. Are the second order partial derivatives of ff at (0,0)(0,0) continuous? Justify your answer.