# Points at which partial derivatives exist and are continuous

Given $E = (xy : xy \neq 0 )$ . Let $f : \mathbb R^{2}\longrightarrow \mathbb R$ be defined by $$f(x,y)=\cases{0,& if xy=0\\y\sin\left(\frac{1}{x}\right)+x\sin\left(\frac{1}{y}\right),& otherwise} .$$

Let $S_1$ be set of points in $\mathbb R^{2}$ , where $f_x$ exists and $S_2$ be set of points where $f_y$ exists . Let $E_1$ be points where partial wrt x is continous and $E_2$ be set of points where $f_y$ is continous

I heed to determine what $S_i$ and $E_i$ are, $i=1,2$

ATTEMPT : I calculated partial derivatives as

$f_x$ = $-(y/x^{2}) \cos(1/x) + x\sin(1/y)$

$f_y = \sin1/x - (x/y^{2} )\cdot\cos (1/y)$

These both partials do not exist at X axis and Y axis and origin. At other points in xy plane they exist. Same for continuity. Is that enough, I'm not confident? Can anyone please elaborate? Thanks

• To type fractions, use the forward slash like p/q for $p/q$ or use \frac{p}{q} for $\frac{p}{q}$. – k.stm Jan 17 '15 at 16:29
• for various values of $m,$ what happens on the lines $y = mx?$ – abel Jan 18 '15 at 5:38
• @abel it ain't exist along that line .how do i proceed ? – Sophie Clad Jan 18 '15 at 5:40
• what does not exist on the lines $y = mx?$ – abel Jan 18 '15 at 5:51
• Both partials . – Sophie Clad Jan 18 '15 at 5:59

let us look at $f_x.$ you have the formula $f_x = -\dfrac{y}{x^2}\cos(1/x) + \sin(1/y)$

it is clear that $f_x$ is continuous everywhere except possibly on the $x$ and $y$ axis. we will look at the axes in turn.

on the $x$-axis, $f$ is identically zero. so $f_x(x, 0) = 0$ for all $x.$

on the $y$-axis, we will use the definition of $f_x$ to compute $f_x.$

for $y \neq 0, f_x(0,y) = \lim_{h \to 0}\dfrac{f(h, y)}{h} =\dfrac{y \sin (1/h) + h\sin (1/y)}{h} = \sin(1/y)+ y\lim_{h \to 0}\dfrac{\sin (1/h)}{h}$ does not exist.

so the set $S_1 = \{(0,y) \colon y \neq 0 \}$

can you do the rest?

• Why is partial wrt x is identically zero in x axis – Sophie Clad Jan 18 '15 at 7:07
• is not $f(x, 0) = 0$ for all $x?$ – abel Jan 18 '15 at 7:10
• In f_x i have calculated we cannot put y=o – Sophie Clad Jan 18 '15 at 7:18
• @SophieClad Use the definition of partial derivative. – Git Gud Jan 18 '15 at 10:21
• @GitGud but when doing it without definition should not produce same results ? – tomb_raider Jan 28 '15 at 12:57