Solution verification: show that the following functions are not differentiable Suppose we have functions,
$f_1: C_1 \to \mathbb{R}$ and $f_2:C_2 \to \mathbb{R},$
where $C_1 = \{x \in \mathbb{R}| x = 1\}, C_2 = \{x \in \mathbb{R}^2| x_2 = 1 - x_1, x_1, x_2 \geq 0\}$
I wish to argue that $\dfrac{df_1(x)}{dx}$ and $\nabla_x f_2(x)$ do not exist.



*

*For the first case,


$\dfrac{df_1(x)}{dx} = \lim\limits_{h \to 0} \dfrac{f_1(x+h) - f_1(x)}{h}$
Should I argue that  the reason for non-differentiability is because:


*

*We do not have another point $x+h$ in the domain of $f_1$


or 


*A point is closed in $\mathbb{R}$, thus the definition of differentiability fails?





*

*For the second case,
Is it possible to show that the following limit fails to exist?
$\lim\limits_{h \to 0} \dfrac{\|f_2(x+h) - f_2(x) - \nabla f_2(x) \cdot h\|}{\|h\|} = 0$


In this case, picking $h$ to form $x+h$ will take us outside of $C_2$, since 
$x_1 + h + x_2 + h = x_1 + h + 1-x_1 + h = 1+2h \implies x+h \notin C_2$.
Is this good enough?

Finally, is there a generalization to this result? When does the derivative (or the gradient) fail to exist due to its domain?
 A: As mentioned in a comment, there's no standard definition for this case. You should not expect people to understand what you mean by differentiability on a non-open set.
Let's use the definition that $T$ is a derivative of $f$ at $x$ if $\|f(y)-f(x)-T(y-x)\|/\|y-x\|\to 0$ whenever $y\to x$ with $y$ in the domain of $f$ but with $y\neq x.$ This limit might be trivial if $x$ is an isolated point, in which case I will say it vacuously converges to $0.$ Then both your functions can have derivatives, it's just that these derivatives aren't unique.
A general condition on $C\subseteq\mathbb R^n$ that a function $C\to\mathbb R$ will never have a unique derivative at $x\in C$ is: there is a proper vector subspace $V\subsetneq \mathbb R^n$ such that $d(y-x,V)/\|y-x\|\to 0$ whenever $y\to x$ with $y\in C\setminus\{x\}.$ Here $d(\cdot,V)$ means distance from the point to the nearest point on $V.$ Given two derivatives $T,T'$ of $f$ at $x,$ they are forced to match on $V,$ and conversely given $T$ you can add any linear map that is zero on $V$ to get a different derivative. You could also phrase this condition by saying that the contingent cone does not span the ambient space.
