1
$\begingroup$

Given a function $F:\mathbb{R}^2\to\mathbb{R}$, by the implicit function theorem, the relation $F(x,y) = 0$, defines $y$ implicitly as function of $x$ in a neirbourhood of $(x_0,y_0)$, provided $F(x_0,y_0)=0$ and $\partial F/\partial y (x_0,y_0)\neq 0$.

Moreover, if the implicit function is $\phi$ then for that neirbourhood holds:

$$\frac {d \phi }{d x} = -\frac {\partial F / \partial x} {\partial F / \partial y}.$$

This result can be obtained differentiating the relation $F(x,\phi(x)) = 0$ with respect to x

$$ \frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\,\frac{\partial \phi}{\partial x}=0 $$ and solving for $d\phi/d x$.

Now, if this could be generalized for the case $F:\mathbb{R}^{n+m}\to\mathbb{R}^m$, then it would be a nice direct way of obtaining the formula for all the partial derivatives of the implicit function (i.e its diferential) by simple implicit diferentiation and matrix algebra.

For example, considering $F(X,Y)=0$, where $X\in \mathbb{R}^n$ and $Y\in \mathbb{R}^m$, suppose suited conditions holds for Y being defined implictly as function of X, say $Y = \Phi(X)$. Taking the total derivative of $F(X,Y)=0$ respect to X we obtain:

$$\frac{dF}{dX} = A + B . D\Phi=0$$ where $A$ and $B$ are the submatrices of $DF$ corresponding to variables $X$ and $Y$ respectively. Then solve for $D\Phi$ by matrix algebra: $$D\Phi=-B^{-1}A$$ I think the process is correct, but I'm not not quite sure if the $dF/dX$ notation for the total derivative is right and don't know how to call $A$ and $B$. (Something like $D_XF$ and $D_YF$?)

$\endgroup$

1 Answer 1

2
$\begingroup$

I personally like to use the notation $\dfrac{\partial F}{\partial X}$ for the $X$-portion of the derivative matrix. However, that notation would be reserved for $A$ and I'd write $B=\dfrac{\partial F}{\partial Y}$. You should be writing $g(X) = F(X,\Phi(X))$ and then writing $$Dg = A+B\cdot D\Phi = \frac{\partial F}{\partial X}+\frac{\partial F}{\partial Y}D\Phi\,.$$ As you commented, this gives the usual formula for implicit differentiation that follows from the Implicit Function Theorem, assuming $B=\dfrac{\partial F}{\partial Y}$ is nonsingular.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .