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Let $f:{\mathbb R}^2\rightarrow {\mathbb R}$ be defined as: $$ f(x,y) = \left\{ \begin{array}{ll} \frac{x^3}{x^2 + y^2}, & \ (x,y)\ne(0,0),\\ 0, & \ (x,y)=(0,0).\\ \end{array} \right. $$ Prove that ${\mathrm D}_1f$ and ${\mathrm D}_2f$ are bounded functions in $\mathbb{R}^2$. Also show that $f$ is continuous

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  • $\begingroup$ D_if=limt->0((f(x+te_j)-f(x,y))/t) D_1f=limt->0 ((f((x,y)+t(1,0))-f(x,y))/t) =lim t->0 ((((x+t)^3)/(((x+t)^2)+y^2))-x^3/(x^2 + y^2)))/t = (x^4 + 3xy^2)/((x^2 + y^2)^2) . . D_2f=(-2x^4)/((x^2 + y^2)^2) $\endgroup$
    – user123592
    Jul 11, 2014 at 14:34

1 Answer 1

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  • $f$ is continuous.

Let $b\in\mathbb{R}$. If $x\neq 0$,

$$|f(x,y)|=\left|\frac{x^3}{x^2+y^2}\right|=\frac{\displaystyle\frac{1}{x^2}}{\displaystyle\frac{1}{x^2}}\left|\frac{x^3}{x^2+y^2}\right|=\left|\frac{x}{1+\displaystyle\frac{y^2}{x^2}}\right|=|x|\left|\frac{1}{1+\displaystyle\frac{y^2}{x^2}}\right|\leq |x|$$ So, $0\leq |f(x,y)|\leq |x|$ for all $(x,y)\in\mathbb{R}^2$. It follws that $$\lim_{(x,y)\to(0,b)}f(x,y)=0=f(0,b)$$ and thus $f$ is continuos at $(0,b)$. Since $$\lim_{(x,y)\to(a,b)}f(x,y)=\frac{a^3}{a^2+b^2}=f(a,b)$$ for all $a\neq 0$, we conclude that $f$ is continuous.

  • $D_xf$ is bounded.

Let $b\in\mathbb R$. If $a\neq 0$ the basic rules give us $$D_xf(a,b)=\frac{a^4+3 a^2 b^2}{a^4+2 a^2 b^2+b^4}.$$ and thus

\begin{align*}|D_xf(a,b)| & \leq \left|\frac{a^4}{a^4+2 a^2 b^2+b^4}\right|+\left|\frac{3 a^2 b^2}{a^4+2 a^2 b^2+b^4}\right|\\ \\ & \leq \left|\frac{1}{1+\displaystyle \frac{2 b^2}{a^2}+\displaystyle\frac{b^4}{a^4}}\right|+\frac{3}{2}\left|\frac{1}{\displaystyle\frac{a^4}{2a^2b^2}+1+\displaystyle\frac{b^4}{2a^2b^2}}\right|\\ \\ &\leq1+\frac{3}{2}=\frac{5}{2}. \end{align*}

If $a=0$ the definition of partial derivative give us $$D_x(a,b)=D_xf(0,b)=\lim_{t\to 0}\frac{f(0+t,b)-f(0,b)}{t}=\lim_{t\to 0}\frac{1}{t}\left(\frac{t^3}{t^2+b^2}-0\right)=\lim_{t\to 0}\frac{t^2}{t^2+b^2}=0.$$

Therefore $|D_xf(a,b)|<5$ for all $(a,b)\in \mathbb{R}^2$ and hence $D_xf$ is bounded.

  • $D_yf$ is bounded.

Let $b\in\mathbb R$. If $a\neq 0$ the basic rules give us $$D_yf(a,b)=-\frac{2 a^3 b}{a^4+2 a^2 b^2+b^4}.$$

Therfore,

\begin{align*}|D_yf(a,b)| & = \left|\frac{2 a^3 b}{a^4+2 a^2 b^2+b^4}\right|=\left|\frac{a}{b}\right|\left|\frac{1}{\displaystyle\frac{a^4}{2 a^2 b^2}+1+\displaystyle\frac{b^4}{2 a^2 b^2}}\right|\leq \frac{|a|}{|b|}\leq 1\text{ if }|a|\leq |b|. \end{align*} and

\begin{align*}|D_yf(a,b)| & = \left|\frac{2 a^3 b}{a^4+2 a^2 b^2+b^4}\right|=\left|\frac{2b}{a}\right|\left|\frac{1}{1+\displaystyle\frac{2 b^2}{a^2}+\displaystyle\frac{b^4}{a^4}}\right|\leq 2\frac{|b|}{|a|}<2\text{ if }|b|<|a|. \end{align*}

If $a=0$ the definition of partial derivative give us $$D_y(a,b)=D_xf(0,b)=\lim_{t\to 0}\frac{f(0,t+b)-f(0,b)}{t}=\lim_{t\to 0}\frac{1}{t}\left(0-0\right)=0.$$

Therefore $|D_yf(a,b)|<2$ for all $(a,b)\in \mathbb{R}^2$ and hence $D_yf$ is bounded.

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  • $\begingroup$ D_if=limt->0((f(x+te_j)-f(x,y))/t) D_1f=limt->0 ((f((x,y)+t(1,0))-f(x,y))/t) =lim t->0 ((((x+t)^3)/(((x+t)^2)+y^2))-x^3/(x^2 + y^2)))/t = (x^4 + 3xy^2)/((x^2 + y^2)^2) . . D_2f=(-2x^4)/((x^2 + y^2)^2) $\endgroup$
    – user123592
    Jul 11, 2014 at 14:52

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