I was wondering if there is a way to compute the average change for an implicitly defined function such that as the interval of a change decreases, we converge to the derivative.
For a "standard" function such result is easy to obtain. Suppose we have $y=f(x)$ and want to find the average change in $f(x)$ as $x$ changes from $x_0$ to $x_0+h$. In that case, we have the
$$ \frac{\Delta y}{\Delta x}:= \frac{y(x_0+h)-y(x_0)}{x_0+h-x_0}=\frac{f(x_0+h)-f(x_0)}{(x_0+h) - x_0},$$
and as $h \rightarrow0$ the average change in $f$ converges to $f'$; that is,
$$ lim_{h\rightarrow 0}\frac{\Delta y}{\Delta x} = \frac{d y}{dx} $$
My question: Is there an equivalent expression/relationship for an implicitly defined function?
My Attempt: Let $f(x,y(x))=0$ and suppose I want to compute the average change in $y(x)$ as $x$ varies from $x$ to $x+h$. We know that
$$ f(x+h,y(x+h))=0$$
Subtracting from the above equation $f(x,y(x))$, and performing simple manipulations, I obtain
$$ \frac{f(x+h,y(x+h))-f(x+h,y(x))}{y(x+h)-y(x)}(y(x+h)-y(x)) + \frac{f(x+h,y(x))- f(x,y(x))}{h}h =0$$
Rearranging, I obtain: $$ \frac{y(x+h))-y(x)}{(x+h)-x} = - \frac{\frac{f(x+h,y(x))- f(x,y(x))}{h}}{\frac{f(x+h,y(x+h))-f(x+h,y(x))}{y(x+h)-y(x)}} $$
If I define $\Delta y:= y(x+h)-y(x)$ and $\Delta x:= (x+h)-x$, then I have $$ \frac{\Delta y}{\Delta x} = - \frac{\frac{f(x+h,y(x))- f(x,y(x))}{h}}{\frac{f(x+h,y(x+h))-f(x+h,y(x))}{y(x+h)-y(x)}}, $$
which seems intuitive and seems to capture the average change in y(x) over interval [x,x+h].
Next, I would like to take the limit as $h\rightarrow 0$ and show that the above expression converge to the derivative of $y(x)$ that we obtain by applying the implicit function theorem.
When I take the limit as $h\rightarrow 0$ on the LHS, I can see that the numerator converges to $f_x(x,y)$; but I am not sure how to deal with the denominator. Does the denominator converge to $f_y(x,y)$? Or is there another way to link the average change in $y$ over interval $[x,x+h]$ to $dy/dx$.