Continuity of first partial derivatives for a multivariable piecewise function I have found the partial derivatives $f_x(a,a) = f_y(a,a) = \frac{s''(a)}{2}$ but am having trouble showing they are continuous. It is easy to show that you can find a $\delta > 0$ such that $|(b,b)-(a,a)| < \delta \implies |f_x(b,b) - f_x(a,a)| < \epsilon$ for any $\epsilon > 0$ by the continuity of $s''(x)$, but I having trouble showing $|f_x(c,d) - f_x(a,a)| < \epsilon$ for any $\epsilon > 0$ and $c \neq d$. Do I need to split the continuity argument up into 2 cases for this function or is there a more elegant way to prove it?
 A: For $c \ne d, f_x(c,d)=\dfrac{s(d)-s(c)-s'(c)(d-c)}{(d-c)^2}$.
By MVT, $f_x(c,d)=\dfrac{(s'(\xi)-s'(c))(d-c)}{(d-c)^2}$ for some $\xi$ between c and d, which is the same as $\dfrac{(\xi-c)s''(\eta)}{d-c}$ for some $\eta$ between $\xi$ and c.
$|(c,d)-(a,a)|\lt \delta \implies |c-a|\lt \delta, |d-a|\lt \delta$, and by triangle inequality, $|c-d|\lt 2\delta$. Since $\eta$ is between c and d, we have $|c-\eta|\lt 2\delta, |d-\eta|\lt 2\delta \implies |\eta-a|\lt |c-\eta|+|a-c|\lt3\delta$
Now, $|f_x(c,d)-f_x(a,a)|=|\dfrac{(\xi-c)s''(\eta)}{d-c}-\dfrac{s''(a)}2|\lt |s''(\eta)-\dfrac{s''(a)}2|\le |s''(\eta)-s''(a)|+|s''(a)-\dfrac{s''(a)}2|= |s''(\eta)-s''(a)|+|\dfrac{s''(a)}2|.$ Since $s''(x)$ is continuous, $\forall \epsilon \gt 0$ we can pick a suitable $\delta\gt 0$ such that $|\eta-a|\lt 3\delta \implies |s''(\eta)-s''(a)|\lt \epsilon$. 
Edit for discussion:
If $\dfrac{\xi-c}{d-c}\le \dfrac12$, the continuity is clear. However if $\dfrac{\xi-c}{d-c}\gt \dfrac12$, then $|f_x(c,d)-f_x(a,a)|\gt \dfrac12|s''(\eta)-s''(a)|$, which will be larger than 0 if $s''(\eta)\ne s''(a)$. Then we can pick $0\lt \epsilon_1 \lt \dfrac12|s''(\eta)-s''(a)|$ , and $|f_x(c,d)-f_x(a,a)|$ will always be larger than $\epsilon_1$ whatsoever. 
