I was wondering whether anybody could explain how you derive this implicit differentiation rule: $$\frac{\partial z}{\partial x} = \frac{-\partial f \mathbin{/} \partial x}{\partial f \mathbin{/} \partial z}$$ if you have a function $z$ implicity defined by $f(x,y,z)=0$?
I have read a derivation but can't make any sense of it.
1 Answer
Take $z$ a function of $x,y$. Then it is always true that $$ f(x,y,z(x,y))=0 $$ Its derivative is also $0$. So $$ \frac{\partial f}{\partial x} dx +\frac{\partial f}{\partial x} dy +\frac{\partial f}{\partial z} \cdot \left( \frac{\partial z}{\partial x} dx +\frac{\partial f}{\partial x} dy \right) =0. $$ If we consider the case $dy=0$, $dy \neq 0$, then divide by $dx$ and cross out $dy$, $$ \frac{\partial f}{\partial x} +\frac{\partial f}{\partial z} \frac{\partial z}{\partial x} =0. $$ Or, $$ \frac{\partial z}{\partial x} =-\frac{\partial f / \partial x}{\partial f /\partial z} $$
