Rules for partial derivatives

Question

Im trying to show some calculation rules for partial derivatives. Let $x,y,z$ be variables which are linked by $f(x,y,z) = 0$. Given is a function $w(x,y)$ show that:

1. $\displaystyle \frac{\partial x}{\partial y} \Bigr|_z = \left(\frac{\partial y}{\partial x}\Bigr|_z\right)^{-1}$
2. $\displaystyle (-1) = \left (\frac{\partial x}{\partial y} \Bigr|_z\right) \left(\frac{\partial y}{\partial z}\Bigr|_x\right)\left( \frac{\partial z}{\partial x}\Bigr|_y\right)$
3. $\displaystyle \frac{\partial x}{\partial w} \Bigr|_z = \left(\frac{\partial x}{\partial y}\Bigr|_z \right)\left( \frac{\partial y}{\partial w}\Bigr|_z\right)$

Now I'm confused by the wording of the exercise itself. What does "the variables are linked by $f(x,y,z) = 0$ mean? I guess $x(y,z), y(x,z), z(x,y)$ hence 1. isn't completely trivial? But how do I go on to show such relations? Do I use the definition of the differential quotient or how do I best show this? Im a physicist so please don't crucify me for not knowing this:)

Cheers in advance!

Answer 1

Thermodynamics huh? :)

1. If $f(x,y,z) = 0$ and $z = const$, then $\left .f(x,y,z)\right |_z = g(x,y) = 0$, which in turns means $x = h(y)$ (under the assumption $h$ exists). So if you differentiate it w.r. to $x$ and w.r. to $y$ you'll get $$ \left .\frac {\partial x}{\partial y}\right|_z = h' \\ 1 = \left .h'\frac {\partial y}{\partial x}\right |_z $$ respectively, so $$ \left .\frac {\partial x}{\partial y}\right|_z = \left (\left .\frac {\partial y}{\partial x}\right|_z \right)^{-1} $$

2. Since $f(x,y,z) = 0$ then e.g. $x = x(y,z)$. Find differential of this equation $$ dx = \left .\frac {\partial x}{\partial y} \right |_z dy + \left .\frac {\partial x}{\partial z} \right |_y dz $$ Now, recall that $x = const$, so $dx = 0$, hence $\left .dy \right|_x = \partial y_x$, and $\left . dz\right |_x = \partial z_x$ $$ \left .\frac {\partial x}{\partial y} \right |_z \partial y_x + \left .\frac {\partial x}{\partial z} \right |_y \partial z_x=0 \\ $$ Divide by $\partial z_x$ $$ \left .\frac {\partial x}{\partial y} \right |_z \frac {\partial y_x}{\partial z_x} + \left .\frac {\partial x}{\partial z} \right |_y = \left .\frac {\partial x}{\partial y} \right |_z \left . \frac {\partial y}{\partial z}\right |_x + \left .\frac {\partial x}{\partial z} \right |_y = 0 $$ Multiply it to $\left .\frac {\partial z}{\partial x} \right |_y$ $$ \left .\frac {\partial x}{\partial y} \right |_z \left . \frac {\partial y}{\partial z}\right |_x \left .\frac {\partial z}{\partial x} \right |_y+ 1 = 0 $$

   Update

3. Turned out, this is the easiest one. If $z=const$, then $w = w(x,y)$ and $y = y(x)$. Instead of proving $$ \left . \frac {\partial x}{\partial w} \right |_z = \left . \frac {\partial x}{\partial y} \right |_z \left . \frac {\partial y}{\partial w} \right |_z $$ let's use the fact $\left . \frac {\partial x}{\partial w} \right |_z = \left (\left . \frac {\partial x}{\partial w} \right |_z \right)^{-1}$ and prove $$ \left . \frac {\partial w}{\partial x} \right |_z = \left . \frac {\partial w}{\partial y} \right |_z \left . \frac {\partial y}{\partial x} \right |_z $$ which is simple differentiation by chain rule.