Tutorial note 2 (differentiability and $C^1$ theorem)

Amendment of tutorial note 2:
Two changes have been made:

Example 1. (a) (i) If we let $y=kx$, then for $x\neq0$ we have \[
 f(x,kx)=\frac{k}{1+k^2},
\] therefore different paths produce different limits, thus $\lim_{(x,y)\to (0,0)}f(x,y)$ does not exist.

(ii) Since $f(x,0)=f(0,y)=0$ for $x,y\in \R$, we have \[
f_x(0,0)=\frac{d}{dx}f(x,0)\bigg|_{x=0} = \frac{d}{dx}0\bigg|_{x=0}=0
\] and \[
f_y(0,0)=\frac{d}{dy}f(0,y)\bigg|_{y=0} = \frac{d}{dy}0\bigg|_{y=0}=0.
\] (iii) For $(x,y)\neq (0,0)$, we have \[
f_x=\frac{y^3-x^2y}{(x^2+y^2)^2} \quad\text{and}\quad f_y=\frac{x^3-xy^2}{(x^2+y^2)^2},
\] and thus $f_x(0,y)=1/y\to \infty$ as $y\to 0$ and $f_y(x,0)=1/x\to \infty$ as $x\to 0$, so $f_x,f_y$ are both not continuous at $(0,0)$.

(b) (i) Since for $(x,y)\neq (0,0)$ we have \[
|f(x,y)|=\left|\frac{x^2}{x^2+y^2}\right|\cdot |y|\leq |y|\leq \sqrt{x^2+y^2}=\|(x,y)\|,
\] thus $\lim_{(x,y)\to (0,0)}f(x,y)=0$.

(ii) We still have $f(x,0)=0$ and $f(0,y)=0$ for each $x,y\in \R$, therefore \[f_x(0,0)=f_y(0,0)=0\] for the same reason as in (a) (ii).

(iii) For $(x,y)\neq (0,0)$ we have \[
f_x=\frac{2xy^3}{(x^2+y^2)^2}\quad\text{and}\quad f_y=\frac{x^4-x^2y^2}{(x^2+y^2)^2}
\] and thus by polar coordinate $(x,y)=(r\cos\theta,r\sin\theta)$, where $\theta=\theta(r)$, we have \[
f_x=2\cos \theta \sin^3\theta\quad\text{and}\quad f_y=\cos^4\theta-\cos^2\theta\sin^2\theta
\] thus different choices of $\theta(r)$ (i.e., different paths) will product different limits, $f_x,f_y$ are not continuous at $(0,0)$.

(c) (i) By polar coordinate we have $f = r^3\cos^2\theta\sin^2\theta$, therefore $f\to 0$ as $(x,y)\to (0,0)$.

(ii) We still have $f(x,0)=f(0,y)=0$ for each $x,y\in \R$, therefore \[
f_x(0,0)=f_y(0,0)=0
\] as in part (a) (ii).

(iii) For $(x,y)\neq (0,0)$ we have \[
f_x=\frac{x^3y^2+2xy^4}{(x^2+y^2)^{3/2}}\quad \text{and}\quad f_y=\frac{y^3 x^2+2yx^4}{(x^2+y^2)^{3/2}}
\] both $f_x,f_y\to 0$ as $(x,y)\to (0,0)$, this is obvious by using polar coordinate system

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Example 2. (a) As $\cos$ is an even function, we have $f=\cos(xy)$, thus direct differentiation yields \[
 f_x(a,b)=-b\sin (ab)\quad\text{and}\quad f_y(a,b)=-a\sin (ab).
\] (b) For every $(a,b)\in \R^2$ we have \[
f_x(a,b)=\frac{d}{dx}f(x,b)\bigg|_{x=a}=\frac{d}{dx}(x\sin |b|)\bigg|_{x=a}=\sin |b|
\] Now we compute $f_y(a,b)$.

Case 1.  If $b>0$, then $(x,y)$ near $(a,b)$ implies $y>0$ implies $f(x,y)=x\sin y$, thus direct differentiation gives $f_y(a,b)=a\cos b$.

Case 2. If $b<0$, then similarly we have $(x,y)$ near $(a,b)$ implies $y<0$ implies $f =-x\sin y$, and also direct differentiation gives $f_y(a,b)=-a\cos b$.

Case 3. If $b=0$, then since \[
f_y(a,0)=\frac{d}{dy} f(a,y)\bigg|_{y=0}=\frac{d}{dy}a\sin |y|\bigg|_{y=0} = a\lim_{h\to 0}\frac{\sin |h|}{h},
\] the limit exists only when $a=0$, therefore we conclude \[
f_y(0,0)=0.
\]

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Example 3. We need to find a linear map $T$ such that \[
\frac{f(\vec{x})-f(\vec{0})-T\vec{x}}{\|\vec{x}\|} \to0 \quad \text{as $\vec{x}\to \vec{0}$.}
\] By the definition of differentiability the only candidate of such $T$ is the linear map induced by the Jacobian matrix \[
f'(\vec{0})=\nabla f(\vec{0})=\matrixx{f_x(\vec{0})&f_y(\vec{0})}.
\] Step 1. Since $f(x,0)=f(0,y)=0$ for every $x,y\in \R$, by direct computation we have \[
f_x(0,0) = \frac{d }{dx} f(x,0)\bigg|_{x=0}= \frac{d }{dx} 0\bigg|_{x=0}=0.
\] And similarly $f_y(0,0)=0$.

Step 2. Since $f(0,0)=0$ and $T=f'(0,0)=\matrixx{0&0}$, we have \[
\lim_{\|(x,y)\|\to 0}\frac{f(x,y)-f(0,0)-\matrixx{0&0}[\begin{smallmatrix}x\\y\end{smallmatrix}]}{\|(x,y)\|} =\lim_{\|(x,y)\|\to 0} \frac{|xy|^\alpha \ln(x^2+y^2)}{\sqrt{x^2+y^2}}
\] By polar coordinate the latter expression becomes $2r^{2\alpha -1} \ln r |\cos \theta|^\alpha |\sin\theta|^\alpha$. As $2\alpha > 1$ by the given condition, thus \[
\lim_{\|(x,y)\|\to 0}\frac{f(x,y)-f(0,0)-\matrixx{0&0}[\begin{smallmatrix}x\\y\end{smallmatrix}]}{\|(x,y)\|}=0.\]

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Example 4. By direct computation for every $(x,y)\neq (0,0)$, \[
f_x=-2\alpha \frac{x(x^4+y^4)}{(x^2+y^2)^{\alpha +1}}+\frac{4x^3}{(x^2+y^2)^\alpha}
\] and \[
f_x=-2\alpha \frac{y(x^4+y^4)}{(x^2+y^2)^{\alpha +1}}+\frac{4y^3}{(x^2+y^2)^\alpha},
\] therefore $f$ is $C^1$ at every point $(a,b)\in \R^2\setminus \{(0,0)\}$. It remains to check $f$ is $C^1$ at $(0,0)$, for this, let's compute $f_x(0,0)$ and $f_y(0,0)$. Since $f(x,0)=x^{4-2\alpha}$, $f(0,y)=y^{4-2\alpha}$ and $2\alpha <3$ implies $\boxed{4-2\alpha>1}$, thus we have \[
f_x(0,0) = \frac{d}{dx}x^{4-2\alpha}\bigg|_{x=0} = 0 \quad\text{and}\quad f_y(0,0) = \frac{d}{dy}y^{4-2\alpha}\bigg|_{y=0} = 0.
\] It remains to check $f_x,f_y\to 0$ as $(x,y)\to (0,0)$, this follows easily by using polar coordinate.

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Example 5. By chain rule we have \begin{align*}
(F\circ G)'(2,1)&=F'(G(2,1))G'(2,1)\\
&=\matrixx{-\!\!\!-\!\!\!-\!\!\!-\nabla x-\!\!\!-\!\!\!-\!\!\!-\\
-\!\!\!-\!\!\!-\!\!\!-\nabla y-\!\!\!-\!\!\!-\!\!\!-\\
-\!\!\!-\!\!\!-\!\!\!-\nabla x^2y-\!\!\!-\!\!\!-\!\!\!-
}(3,3)
\matrixx{-\!\!\!-\!\!\!-\!\!\!-\nabla (s+t)-\!\!\!-\!\!\!-\!\!\!-\\
-\!\!\!-\!\!\!-\!\!\!-\nabla (s^2-t^2)-\!\!\!-\!\!\!-\!\!\!-
}(2,1)\\
&=\matrixx{
1&0\\
0&1\\
2xy&x^2
}(3,3)\matrixx{
1&1\\
2s&-2t
}(2,1)\\
&=\matrixx{
1&0\\0&1\\
18&9
}\matrixx{
1&1\\
4&-2
}=\matrixx{
1&1\\
4&-2\\
54&0
}.
\end{align*} Also we have \[
(F\circ G\circ \gamma)'(1)=(F\circ G)'((\gamma(1))\gamma'(1)=(F\circ G)'(2,1)\matrixx{2\\0}=\matrixx{
1&1\\4&-2\\54&0
}\matrixx{
2\\0
}=\matrixx{2\\8\\108}.
\]