Meaningless differentiability of implicit functions Studying calculus and came across this thing: 

When differentiating implicitly, it is assumed that y represents a differentiable function of
  x. If this is not so, then the resulting calculations may be nonsense. For example, if we
  differentiate the equation
  $x^2 + y^2 + 1 = 0 $(13)
  we obtain $2x + 2y\frac{dy}{dx}= 0$ or $\frac{dy}{dx}=-\frac{x}{y}$
  However, this derivative is meaningless because there are no real values of x and y that
  satisfy (13) (why?); and hence (13) does not define any real functions implicitly

They explain why this derivative is meaningless, but I still don't understand it. In what cases derivative can be meaningless?
 A: Since $x^2+y^2 \geq 0$, it follows that $x^2+y^2+1 \geq 1 >0$. Therefore, there are no values that can be plugged in to make it hold. Regarding implicit differentiation, the assumption when you have $F(x,y)=0$ is if $F_y(x,y)\neq 0$, then by the implicit function theorem, we can write $F(x,y(x))=0$. In other words, $y(x)$ is a curve defined implicitly. Then by taking $\frac{d}{dx}$ of both sides, it follows that $F_x(x,y(x)) + F_y(x,y(x))y'(x)=0$ and you can solve for $y'(x).$
A: The same way that solving certain equations can be meaningless. You can "solve" the equation $\sqrt{x}=-1$ by applying algebraic techniques, but then the value of $x$ found this way isn't actually a solution to the original equation. 
You can also find the derivative of $f(x)=\ln(-x^2)$ (as a real valued function) but since the original function does not exist for any particular value of $x\in\mathbb{R}$, anything dependent on $f$, and so its derivative, is also meaningless for the context of the problem.
The point is that algorithms have assumptions. e.g. "Assume there is a solution to this equation," "Assume this function is defined on some interval," etc.  You can apply algorithms blindly, but you can't trust the results unless all assumptions of the algorithm are verified. 
