Is there an easier way to prove a multivariate function is differentiable? $f\colon U \rightarrow \mathbb{R}, (x,y) \mapsto \sqrt{1 - x^2 - y^2}$ where $U = \{(x,y) \mid x^2 + y^2 < 1\}$.
So the definition of differentiability I have is:
$$\lim \limits_{(x,y) \rightarrow (x_0,y_0)} \dfrac{\|f(x,y) - f(x_0,y_0) - [\frac{\partial{f}}{\partial{x}}(x_0,y_0)]\cdot(x - x_0) - [\frac{\partial{f}}{\partial{y}}(x_0,y_0)]\cdot(y - y_0)\|}{\|(x,y) - (x_0,y_0)\|} = 0$$
So I get:
$$\lim \limits_{(x,y) \rightarrow (x_0,y_0)} \dfrac{\|\sqrt{1 - x^2 - y^2} - \sqrt{1 - {x_0}^2 - {y_0}^2} + \frac{x_0(x - x_0)}{\sqrt{1 - {x_0}^2 - {y_0}^2}} + \frac{y_0(y - y_0)}{\sqrt{1 - {x_0}^2 - {y_0}^2}}\|}{\sqrt{(x - x_0)^2 + (y - y_0)^2}} = 0$$
So far it looks to be getting very hairy and I am wondering now whether there is an easier/simpler way than using the definition of the derivative directly, or if I may have missed some algebraic trick from the beginning.
 A: The easiest way to prove that many types of multivariable functions are differentiable is to prove that a few specific multivariable functions are differentiable, then use the fact that a composition of differentiable functions is differentiable. For your specific problem, you could say something like this:


*

*The function $(x,y)\mapsto 1-x^2-y^2$ is differentiable because it is a polynomial function.

*The function $x\mapsto \sqrt{x}$ is differentiable for all $x>0$.

*Therefore the composed function $(x,y)\mapsto\sqrt{1-x^2-y^2}$ is differentiable whenever $1-x^2-y^2>0$, i.e. whenever $x^2+y^2<1$.


Like Ryan Reich said, your method of proof depends on what you already know. If you aren't allowed to use some of the facts in the proof I just gave, you will have to resort to the messy argument you've been working on.
A: It depends what theorems you already know. For instance, on the domain given, your function is the composition of two differentiable functions, hence differentiable. (The important point is that polynomials are everywhere differentiable and $\sqrt t$ is for $t>0$.) If you don't have these theorems, then no, you just have to suffer with the definition.
