5
$\begingroup$

Suppose I have a function $ f(x, y, g(x, y)) $

How would I express $ \frac{\partial f}{\partial x} $? Using the chain rule, you'd naturally come up with $ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial g} \frac{\partial g}{\partial x} $, except in this expression, $ \frac{\partial f}{\partial x} $ is really only the partial derivative of $f$ with respect to that one parameter, and not $x$. So, my question is, what notation would I use to show this differentiation that is less ambiguous and meaningless than $ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial g} \frac{\partial g}{\partial x} $?

$\endgroup$

1 Answer 1

2
$\begingroup$

You have a function $f(x,y,z)$ presumably, and then you take a composition $h(x,y) = f(x,y,g(x,y))$. The chain rule here is

$$ \frac{\partial h}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial g}{\partial x} $$

and similarly for $y$. You should verify that on your own, and check a couple examples to convince yourself.

$\endgroup$
2
  • $\begingroup$ I don't think you understood what I was trying to ask, but your answer explained what I wanted to know anyway. My error was in thinking that the $ \partial f / \partial x $ on the left side which held only $y$ fixed (which you wrote as $ \partial h / \partial x $ to remove ambiguity) represented the same thing as the $ \partial f / \partial x $ on the right side, which held both $y$ and $g(x, y)$ fixed. $\endgroup$
    – glaba
    Dec 17, 2014 at 5:36
  • 1
    $\begingroup$ Introducing the extra letters h and z made it clear what exactly was being held constant. $\endgroup$
    – glaba
    Dec 17, 2014 at 5:38

You must log in to answer this question.

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