# Implicit differentiation with multiple variables help

I need some help with some implicit differentiation. If this is a trivial question I can read about in a book then can you please recommend which book? I have looked up basic calculus books like Thomas and Finney, Swokowski etc but they don't seem to have good coverage of implicit differentiation with multiple variables.

I have a set up where $x$ is a choice variable that someone gets to pick. For each choice of $x$ the following non-linear system of equations gets solved for $y_1$ and $y_2$ . \begin{eqnarray*} F(x,y_1,y_2)=0 \newline G(x,y_1,y_2)=0 \end{eqnarray*}

and spits out a unique $(x,y_1(x),y_2(x))$ . This $(x,y_1(x),y_2(x))$ is then used to evaluate a third expression $H(x,y_1,y_2)$ . Now I am writing $y_1(x)$ and $y_2(x)$ but $y_1$ and $y_2$ cannot be expressed in terms of elementary functions of $x$ since the above system of equations cannot be solved analytically.

I am sort of looking for an $(x,y_1,y_2)$ where $\frac{dH(x,y_1(x),y_2(x))}{dx}=0$ but what I really want in the expression $\frac{dH(x,y_1(x),y_2(x))}{dx}$ which I need to plug somewhere else.

I had initially thought that I would assume that $y_1$ and $y_2$ are functions of $x$ and differentiate the system of equations w.r.t. $x$ to solve for $\frac{dy_1}{dx}$ and $\frac{dy_2}{dx}$ \begin{eqnarray*} F_x+F_{y_1}\frac{dy_1}{d_x}+F_{y_2}\frac{dy_2}{dx}=0 \newline G_x+G_{y_1}\frac{dy_1}{d_x}+G_{y_2}\frac{dy_2}{dx}=0 \end{eqnarray*}

and substitute in $\frac{dH(x,y_1(x),y_2(x))}{dx}$ but I am confused if I also need to do cross partials and consider $y_1$ to be a function of $y_2$ and vice-versa.

Any help understanding the math here at a conceptual level and making progress on this problem will be greatly appreciated.

-
Do you want to calculate $dH/dx$ or to equate it to zero and solve for $x$? The optimal process may depend on the objective. To answer your question: The equations $F=G=0$ determine $y_1$ and $y_2$ as functions of single variable $x$. So the only variable in which you can differentiate them is $x$. The idea of solving a 2 by 2 system to get those derivatives is right on target. – user31373 Aug 14 '12 at 5:30
Hi Leonid, I just want to calculate $\frac{dH}{dx}$. Ok so I am doing the right thing. sweet. I was just so confused. Thanks. – Amatya Aug 14 '12 at 5:43