# Functions of two variables and small increments.

#### bluntpencil

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
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 $$\displaystyle \frac{\partial f}{\partial y}= \frac{df}{dg}\frac{\partial g}{\partial y}$$.

• bluntpencil

#### bluntpencil

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)