Functions of two variables and small increments.

Aug 2008
6
0
Okay,

I understand that if a differentiable function \(\displaystyle z=f(g(x,y))\) and x, y are the independent variables, then

\(\displaystyle \frac{\partial{g}}{\partial{x}} = \lim_{\delta{x}\to{0}}\left(\frac{\delta{g}}{\delta{x}}\right)\) and \(\displaystyle \frac{\partial{f}}{\partial{g}} = \lim_{\delta{g}\to{0}}\left(\frac{\delta{f}}{\delta{g}}\right)\), so \(\displaystyle \frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{g}}.\frac{\partial{g}}{\partial{x}}\), keeping y constant

and

\(\displaystyle \frac{\partial{g}}{\partial{y}} = \lim_{\delta{y}\to{0}}\left(\frac{\delta{g}}{\delta{y}}\right)\) and \(\displaystyle \frac{\partial{f}}{\partial{g}} = \lim_{\delta{g}\to{0}}\left(\frac{\delta{f}}{\delta{g}}\right)\), so \(\displaystyle \frac{\partial{f}}{\partial{y}} = \frac{\partial{f}}{\partial{g}}.\frac{\partial{g}}{\partial{y}}\), keeping x constant

now, if I let \(\displaystyle u=g(x, y)\) can someone show/explain using the above method what \(\displaystyle \frac{\partial{z}}{\partial{u}}\) would be?.

Thanks.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Okay,

I understand that if a differentiable function \(\displaystyle z=f(g(x,y))\) and x, y are the independent variables, then

\(\displaystyle \frac{\partial{g}}{\partial{x}} = \lim_{\delta{x}\to{0}}\left(\frac{\delta{g}}{\delta{x}}\right)\) and \(\displaystyle \frac{\partial{f}}{\partial{g}} = \lim_{\delta{g}\to{0}}\left(\frac{\delta{f}}{\delta{g}}\right)\), so \(\displaystyle \frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{g}}.\frac{\partial{g}}{\partial{x}}\), keeping y constant

and

\(\displaystyle \frac{\partial{g}}{\partial{y}} = \lim_{\delta{y}\to{0}}\left(\frac{\delta{g}}{\delta{y}}\right)\) and \(\displaystyle \frac{\partial{f}}{\partial{g}} = \lim_{\delta{g}\to{0}}\left(\frac{\delta{f}}{\delta{g}}\right)\), so \(\displaystyle \frac{\partial{f}}{\partial{y}} = \frac{\partial{f}}{\partial{g}}.\frac{\partial{g}}{\partial{y}}\), keeping x constant

now, if I let \(\displaystyle u=g(x, y)\) can someone show/explain using the above method what \(\displaystyle \frac{\partial{z}}{\partial{u}}\) would be?.

Thanks.
If, as you say, z= f(g(x,y)) and u= g(x,y), then you have z= f(u) and there are no "partial" derivatives. You should be writing \(\displaystyle \frac{dz}{du}\).

\(\displaystyle \frac{dz}{du}= \frac{dz}{du}\) since "u" and "g" are equal- they are just different ways of writing the same thing.

In fact, you should not be writing "\(\displaystyle \frac{\partial f}{\partial g}\)" because f is a function of the single variable g.

The correct chain rules are \(\displaystyle \frac{\partial f}{\partial x}= \frac{df}{dg}\frac{\partial g}{\partial x}\) and \(\displaystyle \frac{\partial f}{\partial y}= \frac{df}{dg}\frac{\partial g}{\partial y}\).
 
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Aug 2008
6
0
I stand corrected. I never could get the hang of the notation, until now. This has helped me a lot in rates of change and change of variable problems.

Much appreciated. (Happy)