- #1

- 3

- 0

(del x/ del y)_z = part of x with respect to y, hold z constant

I dont know why is it negative 1?

del x/ del y)_z * del y/ del z)_x * del z/ del x)_y = -1

- Thread starter gloryofgreece
- Start date

- #1

- 3

- 0

(del x/ del y)_z = part of x with respect to y, hold z constant

I dont know why is it negative 1?

del x/ del y)_z * del y/ del z)_x * del z/ del x)_y = -1

- #2

- 1,015

- 3

[tex]\mathrm{d}z=\left.\frac{\partial z}{\partial x}\right)_y\mathrm{d}x+\left.\frac{\partial z}{\partial y}\right)_x\mathrm{d}y[/tex]

Now to get [itex]\left.\frac{\partial x}{\partial y}\right)_z[/itex] from the above equation just find the fraction [itex]\frac{\mathrm{d}x}{\mathrm{d}y}[/itex] under the condition that [itex]\mathrm{d}z=0[/itex], i.e.

[tex]\left.\frac{\partial x}{\partial y}\right)_z=\left.\frac{\mathrm{d} x}{\mathrm{d} y}\right)_{\mathrm{d}z=0}=-\frac{\left.\partial z/\partial y\right)_x}{\left.\partial z/\partial x\right)_y}[/tex]

The minus sign comes from rearranging the first equation.

Also

[tex]\left.\frac{\partial z}{\partial y}\right)_x=\frac{1}{\left.\frac{\partial y}{\partial z}\right)_x}[/tex]

or

[tex]\left.\frac{\partial z}{\partial y}\right)_x\left.\frac{\partial y}{\partial z}\right)_x=1[/tex]

The apparent cancellation is only possible since both derivatives have the same variables kept constant!

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