I have come across following:
$$\frac{dI_{(V)}}{dV} = - \frac{\frac{\partial f_{(V,I)}}{\partial V}}{\frac{\partial f_{(V,I)}}{\partial I}}$$
where:
- $I_{(V)}$ is transcendental function (a function with both $I$ and $V$ as variables)
- and $f_{(V,I)} = I_{(V)} - I = 0$, which is often used for instance in Newton-Raphson method.
I can provide the transcendental function in question:
$$I = I_{ph} - I_s(exp\left(\frac{V+IR_s}{v_T}\right)-1)-\frac{V+IR_s}{R_{sh}}-a(V+IR_s)\left(1-\frac{V+IR_s}{V_{Br}}\right)^{-n}$$
and for completeness, the $f_{(V,I)}$ is then:
\begin{align} f_{(V,I)} & = I_{ph} - I_s(exp\left(\frac{V+IR_s}{v_T}\right)-1)-\frac{V+IR_s}{R_{sh}}-a(V+IR_s)\left(1-\frac{V+IR_s}{V_{Br}}\right)^{-n} - I \\ & = 0 \end{align}
QUESTION:
Why the total derivative $\frac{dI_{(V)}}{dV}$ equals to ratio of partial derivatives $- \frac{\frac{\partial f_{(V,I)}}{\partial V}}{\frac{\partial f_{(V,I)}}{\partial I}}$ ?
How do I show this holds in general case ?