# Implicit derivative?

If we have function $y=L(x_1,x_2,x_3,...,x_n)$, and function $z=R(x_1,x_2,x_3,...,x_n)$. How to compute the derivative $\frac{dy}{dz}$?

Shall I do $\frac{dy}{dz} = \sup_{g\in \Re^n}\frac{\bigtriangledown_x L \cdot g}{\bigtriangledown_x R \cdot g}$?

Is there any mathematical term associated with this kind of derivatives?

## 2 Answers

There is no reason to expect that such a thing as $dy/dz$ exists. If you rephrase everything in terms of differentials, you have $$dy=\sum_{i=1}^n\frac{\partial L}{\partial x_i}dx_i,\quad dz=\sum_{i=1}^n\frac{\partial R}{\partial x_i}dx_i$$ where $dx_1,\ldots,dx_n$ are linearly independent, and so you find $dy=a\, dz$ if and only if $\nabla L=a\nabla R$. But normally, the two gradient won't be parallel, so you cannot define $dy/dz$. You can do so in any given direction, though: Your fraction $(\nabla L\cdot g)/(\nabla R\cdot g)$ is a perfectly good expression for $dy/dz$ as measured in the direction given by $g$.

• Agree. The sup must lead to infinity if the two gradients are not parallel. Thanks. – pacificmoth Apr 25 '10 at 19:34

In general you cannot even define $y(z)$ given generic functions $R$ and $L$.

If instead there exists $f$ such that $y=f(z)$ is satisfied for any point $\mathbf{x}=(x_1,...,x_n)$ then you have $$L(\mathbf{x})=y=f(z)=f(R(\mathbf{x}))$$ which is a nontrivial relation between the functions $R$ and $L$. This relation in particular implies $$\nabla L(\mathbf{x})=\nabla R(\mathbf{x}) f'(R(\mathbf{x}))$$ that means the gradients are parallel in any point. This relation is in fact an array of relations which uniquely define $f'=\frac{dy}{dz}$ given $L$ and $R$.