# Differentiability of $f(x,y)=\frac{\sin^4(x)\sin^4(y)}{x^2+y^2}$ extended by $f(0,0)=0$

$f:\mathbb{R}^2\rightarrow \mathbb{R}, (x,y)\mapsto \begin{cases} \frac{\sin^4(x)\sin^4(y)}{x^2+y^2} &\text{if$(x,y) \neq (0,0)$}\\0&\text{if$ (x,y)=(0,0)$}\end{cases}$

(i) Show that $f$ is differentiable on its domain of definition and calculate $f'(0,0)$

(ii) Determine the Gradient in $P=(\frac{\pi}{2},\frac{\pi}{2})$

(iii) Find a direction $\vec {v}$, whose Directional derivative $\frac{\partial f}{\partial \vec {v} }$ in $P$ is $0$

For (i) my idea was, to show that its partial derivative is continuous:

$f'(x,y) = \frac{2 \sin^3(x) \cos^4(y) (2 (x^2+y^2) \cos(x)-x \sin(x))}{(x^2+y^2)^2} \approx \frac{2 x^3 y^4 (2 (x^2+y^2) x-x^2)}{(x^2+y^2)^2}$ for small $x$ and $y$ is this correct? How to go on?

To (iii)

$\langle \nabla f (\frac{\pi}{2},\frac{\pi}{2}), \vec {v} \rangle = \langle \begin {pmatrix} -\frac{\pi}{2} \\ -\frac{\pi}{2} \end{pmatrix}, \begin {pmatrix} v_1 \\ v_2 \end{pmatrix} \rangle = -\frac{\pi}{2}(v_1+v_2)=0 \Rightarrow v_1=v_2$ So $\vec {v}=(1,1)^T$ would be a solution. Is this correct?

• In (i): "...and f'$(0,0)$" what? Commented Nov 16, 2014 at 23:16

• Why is it allowed, to disregard $y$ and just consider $x$ using L'Hôpital's rule? Commented Nov 17, 2014 at 1:03