Functions of two variables and small increments.

**bluntpencil** · May 18th 2010, 02:41 PM

Okay,

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

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

and

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

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

Thanks.

**HallsofIvy** · May 19th 2010, 04:34 AM

Originally Posted by bluntpencil

Okay,

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

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

and

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

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

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

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

The correct chain rules are $\frac{\partial f}{\partial x}= \frac{df}{dg}\frac{\partial g}{\partial x}$ and $\frac{\partial f}{\partial y}= \frac{df}{dg}\frac{\partial g}{\partial y}$ .

**bluntpencil** · May 19th 2010, 01:28 PM

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.

Thread: Functions of two variables and small increments.

LinkBack

Thread Tools

Display

Functions of two variables and small increments.

Similar Math Help Forum Discussions

random variables separable into 2 functions integrated over the same variables

Increments of area dxdy

salary increments

Small Increments 1

Constructing independent increments

Search Tags