constrained partial derivative identity

**FGT12** · March 6th 2011, 03:08 AM

if $f(x,y,z)=0$ then,
$(\frac{\partial x}{\partial y})_z (\frac{\partial y}{\partial z})_x (\frac{\partial z}{\partial x})_y = -1$

I tried expressing $\frac{\partial f}{\partial x}$ with z constant but this didnt work

Im lost. Help much appreciated

**HallsofIvy** · March 6th 2011, 03:22 AM

It should! With z constant and x a function of y, we can think of f as a function of y only:
$\frac{df}{dy}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial y}+ \frac{\partial f}{\partial y}= 0$
so
$\left(\frac{\partial x}{\partial y}\right)_z= -\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$

Now do the same for the other two

**FGT12** · March 6th 2011, 03:31 AM

Why have you written $\frac{df}{dy}$ should it not be $\frac{\partial f}{\partial y}$ as $z=g(x,y)$

This is where my confusion arises

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