It is well know that in order to calculate the algorithmic error of a function one can use backward analysis with using the visual representation of a graph with nodes the $i-$th step of the algorithm and as weight the amplification coefficient of the operation, which can be calculated via Taylor's formulas for istance.
In the case of $f(x) = \frac{e^x -1}{x}$ I guess that I'd like to give the following operations
$(op)_1 \bullet x \to e^x$
$(op)_2 \bullet e^x \to e^x-1$
$(op)_2\bullet e^x-1 \to \frac{e^x -1}{x} = f(x)$
Making the diagram I see that the algorithmic error in the last step should be a local error, $\delta_3$, plus the amplification coefficient of the division of $e^x-1$ and $x$, which should be 1 on the arc coming from the arc of $e^x-1$ and $-1$ on the arc coming from $x$, accordin to the amplification coefficients of the usual operations. What worries me is that if I make the calculus explictly I don't find that $1$ is the amplification coefficient since which in general, with a given $f(\textbf{x})$ I would find as $c_{x_i} = \frac{x_i}{f(\textbf{x})}\cdot \frac{\partial f(\textbf{x})}{\partial x_i}$, since the variables here ($e^x-1$ and $x$) are not "independent" (if makes any sense).
I believe there is an error in my reasoning but I don't understand where, any help understanding this would be appreciated. Also $e^x$ is not a rational function hence I don't know whether the graph can be done in the exact same way, what if there is a library function $\operatorname{Exp}$ such that $\operatorname{Exp}(x) = e^x(1+\epsilon)$, does this would change the answer?
Thanks in advance.