Applications of Partial Derivatives on $f(x, y) = \sqrt{\frac{1}{|1-x^2 - y^2|}}$
a) Find the domain of the function.
I think the domain should be real numbers without a circle with radius 1, and beginning at [0, 0]. Basically $x^2 + y^2 \ne 1$. Is that correct?
b) Prove whether $f(x, y)$ is totally differentiable on [1,1]. If it is, write it's total derivative.
So I know that a function has a total derivative (is totally differentiable) at a point if there exists a linear transformation $L: \mathbb{R}^m \rightarrow \mathbb{R}^n $ such that $\lim \limits_{||h|| \to 0} \frac{||f(a+h)-f(a)-L(h)||}{||h||} = 0$.
I know that if $f(x,y)$ is differentiable at $a$ then the total derivative would just be $D_f(a)(h) =\frac{\partial f}{\partial x}(a)\cdot h_1 + \frac{\partial f}{\partial y}(a)\cdot h_2$
But how would I go about proving that the function is differentiable at $[1,1]$?
c) Approximate at $f(0,98; 1,01)$
I know that if $f(x,y)$ is differentiable at $a$, then in has local approximations near $a$: $f(a+h) = f(a) + D_f(a)(h) + o(||h||), ||h|| \to 0$
But I'm not sure whether this is the correct equation to apply.
(a) The domain is $U = \{(x,y) \in \mathbb{R}^2: x^2 + y^2 \neq 1\}$, as you said.
(b) If you know the theorem which states that if the partial derivatives $\partial_1f$ and $\partial_2f$ are continuous at $(1,1)$, then $f$ is differentiable at $(1,1)$, proving differentiability is very easy; just compute the partials and show they are continuous.
If you don't know that theorem, then yes, you are right that $Df(a)(h)$ is indeed given by that formula. So, before finding a candidate linear transformation $L$, you need to compute the partials at $(1,1)$. Note that to compute $(\partial_1f)(1,1)$, you have to consider the function $g(x) = f(x,1)$, and compute $g'(1)$. This is easy: \begin{equation} g(x) = \dfrac{1}{x} \end{equation} hence, $g'(1) = -1$. So, $\partial_1f(1,1) = -1$. Perform a similar computation for $\partial_2f(1,1)$. Then, do some tough algebra and try to show that $L$ defined by \begin{equation} L(h,k) = \partial_1f(1,1) \cdot h + \partial_2f(1,1) \cdot k \end{equation} satisfies the difference quotient limit you wrote down.
(c) Yes, the equation you wrote down is correct; choose $a = (1,1)$, and $h= (-0.02, 0.01)$, and use the linear approximation \begin{equation} f(a+h) \approx f(a) + Df(a)(h). \end{equation}