# Study directional derivative for $f(x,y)$ at $(x,y)=(0,1)$

Let $$f(x,y)=\begin{cases}\dfrac{{(y-1)}^2\sin{(5x)}}{{(y-1)}^2+x^2}&\text{if}&(x,y)\neq(0,1)\\0&\text{if}&(x,y)=(0,1).\end{cases}$$ Study the existence of the directional derivative for $(x,y)=(0,1)$.

Let $\check v=(a,b)$.

$$\underset{\text{with }a^2+b^2=1}{f'\big((0,1);(a,b)\big)}\underbrace{=}_{\text{If exists}}\displaystyle\lim_{h\to0}{\dfrac{f(ha,hb+1)-\overbrace{f(0,1)}^0}h}=\displaystyle\lim_{h\to0}{\dfrac{h^2b^2\sin{(5ha)}}{h^2b^2+h^2a^2}\dfrac1h}=\displaystyle\lim_{h\to0}{\dfrac{h^2b^2\sin{(5ha)}}{h^3\underbrace{\left(a^2+b^2\right)}_1}}=\displaystyle\lim_{h\to0}{\dfrac{b^2\sin{(5ha)}}{h}}=\displaystyle\lim_{h\to0}{\dfrac{5ab^2\sin{(5ha)}}{5ah}}=\displaystyle\lim_{h\to0}{\left(5ab^2\right)}=5ab^2\;\therefore\;\boxed{\exists f'\big((0,1);(a,b)\big)\;\forall\check v}.$$

Is that correct?

Thanks!

• Yes, this is just fine. Well done. – Ted Shifrin Aug 22 '18 at 1:32