Prove that a function is differentiable using the limit definition Use the definition of the derivative to prove that $f(x,y)=xy$ is differentiable. So we have: $$\lim_{h \to 0} \frac{||f(x_0 + h) - f(x_0) - J(h)||}{||h||} = 0$$ We find the partial derivatives which are $f_x = y$ and $f_y = x$. We plug them into the definition: $$\lim_{h \to 0} \frac{||f(x_0 + h) - f(x_0) - yh - xh||}{||h||} = 0$$ I'm not sure what to do from here. So do we calculate the norm of the numerator and denominator now?
 A: $\displaystyle\lim_{h \to 0} \frac{\|(x+h)(y+h) - xy - yh - xh\|}{\|h\|} =$
$\displaystyle\lim_{h \to 0} \frac{\|xy + xh + hy + h^2 - xy - yh - xh\|}{\|h\|} =$
$\displaystyle\lim_{h \to 0} \frac{\| h^2\|}{\|h\|} = $
$\displaystyle\lim_{h \to 0} \|h\| = 0$
A: Remember $x_0$ and $h$ are vectors. Let's write with bar notation to make it visually different. Here, $\bar{x}_0=(x,y)$. Let $\bar{h}=(\Delta x,\Delta y)$. Then $\bar{x}_0+h=(x+\Delta x,y+\Delta y)$. (We cannot simply assume that $\bar{h}$ has the from $(h,h)$. ) We can use the function $J(\Delta x,\Delta y)=\Delta x\cdot y+\Delta y\cdot x$
Now, 
$\displaystyle\lim_{h \to 0} \frac{\|f(\bar{x}_0+\bar{h})- f(\bar{x}_0) - J(\bar{h}) \|}{\|\bar{h}\|} =$
$\displaystyle\lim_{h \to 0} \frac{\|(x+\Delta x)(y+\Delta y) - \Delta x\cdot \Delta y - \Delta x\cdot y- \Delta y\cdot x\|}{\|\bar{h}\|} =$
$\displaystyle\lim_{h \to 0} \frac{\|\Delta x\cdot \Delta y\|}{\sqrt{(\Delta x)^2+(\Delta y)^2}}$
Note that $$\frac{\|\Delta x\cdot \Delta y\|}{\sqrt{(\Delta x)^2+(\Delta y)^2}}\leq \frac{\|\Delta x\cdot \Delta y\|}{\sqrt{(\Delta x)^2}}\leq \frac{\|\Delta x \| \|\Delta y\|}{\|\Delta x\|}=\|\Delta y\|$$ provided $\Delta x\neq0$. Similarly the ratio is less than or equal to $\|\Delta x\|$ provided $\Delta y\neq0$.
Now $\|\Delta x\|\to 0$ and $\|\Delta y\|\to 0$ as $\bar{h}\to0$. This ensures that $\frac{\|\Delta x\cdot \Delta y\|}{\sqrt{(\Delta x)^2+(\Delta y)^2}}\to 0$ as $\bar{h}\to 0$ by application of squeeze theorem on $0\leq\frac{\|\Delta x\cdot \Delta y\|}{\sqrt{(\Delta x)^2+(\Delta y)^2}}\leq \max\{\|\Delta x\|,\|\Delta y\|\}$.
I hope the answer is complete now. Please correct me otherwise.
