Direct proof for differentiability of $\sin(x+y)$ I've been triying expand the functions like this:
$$\lim_{(x,y)\rightarrow(x_0,y_0)}\frac{|\sin(x+y)-\sin(x_0+y_0)-(\cos(x_0+y_0),\cos(x_0+y_0))\cdot(x-x_0,y-y_0)|}{||(x-x_0,y-y_0)||}\\
\lim_{(x,y)\rightarrow(x_0,y_0)}\frac{|\sin(x)\cos(y)+\sin(y)\cos(x)-\sin(x_0)\cos(y_0)-\sin(y_0)\cos(x_0)-\left[\cos(x_0)\cos(y_0)-\sin(x_0)\sin(y_0)\right]\left[(x-x_0)+(y-y_0)\right]|}{||(x-x_0,y-y_0)||}$$
But I don't know how to continue, \? Any hint to prove that this limit is actually zero?. 
 A: Once you have proved that both $\sin(z)$ and $x + y$ are differentiable , then the differentiability of $\sin(x + y)$ follows from a general fact: the composite of two differentiable functions is differentiable.
A: It looks easier in the equivalent form
$$
\lim_{(h,k)\to(0,0)}\frac{f(x+h,y+k)-f(x,y)-\nabla f(x,y)\cdot(h,k)}{\sqrt{h^2+k^2}}=0.
$$
In this form we have to calculate the limit of 
$$
\frac{\sin(x+y+h+k)-\sin(x+y)-\cos(x+y)(h+k)}{\sqrt{h^2+k^2}}.
$$
Calling $a=x+y$ and $t=h+k$ it becomes
$$
\frac{\sin(a+t)-\sin a-t\cos a}{t}\cdot\frac{h+k}{\sqrt{h^2+k^2}}.
$$
As $(h,k)\to 0$ we have that $t\to 0$ as well. It is now left to show that


*

*the first fraction goes to zero (can show directly or appeal to differentiability of $\sin$),

*the second fraction is bounded (e.g. by $2$).

A: $$\dfrac{\partial}{\partial x} \sin(x+y) =\lim\limits_{\Delta x\to 0}\dfrac{\sin(x+y+\Delta x)-\sin(x+y)}{\Delta x} =\lim\limits_{\Delta x\to 0}\dfrac{2\cos\dfrac{2x+2y+\Delta x}2\sin\dfrac{\Delta x}2}{\Delta x}$$
$$=\lim\limits_{\Delta x\to 0}\cos\dfrac{2x+2y+\Delta x}2
\lim\limits_{\Delta x\to 0}\dfrac{\sin\dfrac{\Delta x}2}{\dfrac{\Delta x}2} = \cos(x+y)\quad \forall (x,y)\in\mathbb R^2,$$ 
and similarly for $\dfrac{\partial}{\partial y}.$
If $\Delta x$ and $\Delta y$ changes by the law 
$$\Delta x = \Delta l\cos \varphi,\quad \Delta y = \Delta l\sin\varphi,$$ then
$$\dfrac{\partial}{\partial l} \sin(x+y) =\lim\limits_{\Delta l\to 0}
\dfrac{\sin(x+y+\Delta x+\Delta y)-\sin(x+y)}{\Delta l}$$
$$ = \lim\limits_{\Delta l\to 0}
\dfrac{\sin(x+y+\Delta x+\Delta y)-\sin(x+y+\Delta y)}{\Delta l}
+\lim\limits_{\Delta l\to 0}
\dfrac{\sin(x+y+\Delta y)-\sin(x+y)}{\Delta l} $$
$$ = \cos\varphi\lim\limits_{\Delta l\to 0}
\dfrac{\sin(x+y+\Delta l\cos\varphi+\Delta l\sin\varphi)-\sin(x+y+\Delta l\sin\varphi)}{\Delta l\cos\varphi}$$
$$+\sin\varphi \lim\limits_{\Delta l\to 0}
\dfrac{\sin(x+y+\Delta l\sin\varphi)-\sin(x+y)}{\Delta l\sin\varphi} $$
$$=\dfrac{\partial \sin(x+y)}{\partial x}\cos\varphi
+ \dfrac{\partial \sin(x+y)}{\partial y}\sin\varphi.$$
