Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to show that the function:

$$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$$

is a $C^1$-function.

$$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$$

$$ \frac{\partial f}{\partial y}(x,y)=-\frac{\partial f}{\partial y}(y,x)=... $$

So I just have to show that:

$$ \frac{\partial f}{\partial x}(x,y)\rightarrow_{(0,0)}0$$

When $y\geq0$ :

$$ -\frac{y^3}{6(x^2+y^2)}+\frac{x^2y}{x^2+y^2}-\frac{x^4y}{4!(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}$$

When $y<0$ :

$$ -\frac{y^3}{6(x^2+y^2)}+\frac{y^5}{5!(x^2+y^2)}+\frac{x^2y}{2(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}-\frac{yx^4}{4!(x^2+y^2)}$$

So $$ \frac{\sin(y)-y\cos(x)}{x^2+y^2}\rightarrow_{(0,0)}0 $$

How can I directly find an upper bound of $$ \left| \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2} \right|$$ that tends to 0 ?

share|cite|improve this question
up vote 1 down vote accepted

Because of the Taylor expansion $\sin(x) = x - {x^3 /6} + ...$, if $x$ and $y$ are small enough you can write $\sin(x) = x + E(x)$ and $\sin(y) = y + E(y)$, where $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$. So you have $${2x(y\sin(x) - x\sin(y)) \over (x^2 + y^2)^2} = {2x(yx + yE(x) - xy - xE(y)) \over (x^2 + y^2)^2}$$ $$= {2x(yE(x) - xE(y)) \over (x^2 + y^2)^2}$$ Taking absolute values and bounding, this is at most $$= 2|x|{|yE(x)| + |xE(y)| \over (x^2 + y^2)^2}$$ Inserting $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$ this is bounded by $$= 2|x|{|x^3y| + |xy^3| \over (x^2 + y^2)^2}$$ $$= 2|x|{|xy|(x^2 + y^2) \over (x^2 + y^2)^2}$$ $$= 2|x| {|xy| \over x^2 + y^2}$$ I think you can take it from there...

share|cite|improve this answer
So we can write the Taylor expansion of $f(x,y)$ with respect to each variable $x$ and $y$ at the same time to find the limit in $(0,0)$ ? So: $$ \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}=\frac{2x(x(y-y^3/6+o(y^4))-y(x-x^3/6+o‌​(x^4)))}{(x^2+y^2)^2}...\rightarrow 0$$ ? – Chon Mar 19 '12 at 18:34
@Chon Yes, exactly. – Zarrax Mar 19 '12 at 21:20
Thank you for your answer! – Chon Mar 20 '12 at 21:52

A partial answer to at least show that $f$ is differentiable at $(0,0)$. Then it remains to show that its derivative is continuous. I didn't check if that also follows from this bound directly.

Combine $\tan(x)\geq x$ and $\sin(x) \leq x$ for $x \in [0,\pi/2)$ to get

$$ \cos(x) \leq \frac{\sin(x)}{x} \leq 1 $$

for $x \in (-\pi/2,\pi,2)$. Then for $x,y \in (-\pi/2,\pi/2)$

$$ -\frac{y^2}{2} \leq \cos(y)-1 \leq \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \leq 1 - \cos(x) \leq \frac{x^2}{2}. $$

Taking the absolute value:

$$ \left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \right| \leq \frac{\max(x^2,y^2)}{2}\leq \frac{x^2+y^2}{2}. $$

This results in the following estimate of $f$:

$$ \left|\frac{x\sin(y) - y\sin(x)}{x^2+y^2}\right| \leq \left| \frac{xy}{x^2+y^2}\right| \cdot \left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x}\right| \leq \frac{x^2+y^2}{4} $$

This is sufficiently sharp to conclude that $f$ is differentiable in $(0,0)$.

share|cite|improve this answer
Thank you for your answer WimC, even if I already know that $f$ is continuous in $(0,0)$ (it was the subject of my previous post). – Chon Mar 18 '12 at 21:19
@Chon I know, but this shows more than mere continuity. It shows differentiability for one. It remains to show that the derivative is continuous. – WimC Mar 19 '12 at 5:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.