Show that $f(x,y)=\sin (x+y)$ is differentiable in its domain by the definition i.e. prove $\lim_{(x,y) \rightarrow (x_0, y_0)}\frac{|\sin(x+y)-\sin{(x_0+y_0)}-\cos(x_0+y_0)(x-x_0)-\cos(x_0+y_0)(y-y_0)|}{\|(x,y)-(x_0,y_0)\|} = 0$
I can not find a way to compare $|f(x,y)-z|$ with $\|(x,y)-(x_0,y_0)\|$
Consider Taylor series for $\sin\ t$ : $$ \sin\ t=\sin\ t_0 +\cos\ t_0 (t-t_0) -\sin\ t_0 \frac{(t-t_0)^2}{2} +\cdots $$
Hence \begin{align*}&\frac{1}{ |(x,y)-(x_0,y_0)| } \bigg|\sin\ (x+y)-\sin\ (x_0+y_0) -\cos\ (x_0+y_0) \{ x-x_0 +y-y_0\} \bigg| \\&= \frac{1}{ |(x,y)-(x_0,y_0)| }|\sin\ (x_0+y_0)| \frac{|x+y-x_0-y_0|^2}{2} +\cdots \end{align*}
Here $\frac{(a+b)^2}{\sqrt{a^2+b^2}} \leq 2\sqrt{a^2+b^2}$ so that the proof is followed.
