Question

`g(x)=(x-1)/(x+1), x=(z+1)/(1-z)`.Then`(dg)/dz=`?(Express it in z)

`g(x)=(x-1)/(x+1), x=(z+1)/(1-z)`.Then`(dg)/dz=`?(Express it in z)

Answer 1

`\frac{dg}{dz}=1`

Explanation:

`\frac{dg}{dz} = \frac{dg}{dx}\cdot\frac{dx}{dz}`

By quotient rule:

`\frac{dg}{dx}=\frac{(x+1)\cdot 1-(x-1)\cdot 1}{(x+1)^2}=\frac{2}{(x+1)^2}`

`\frac{dx}{dz}=\frac{(1-z)\cdot 1 - (z+1)\cdot (-1)}{(1-z)^2}=\frac{2}{(1-z)^2}`

Therefore,

`\frac{dg}{dz} = \frac{2\cdot 2}{(x+1)^2\cdot(1-z)^2}=\frac{4}{(\frac{z+1}{1-z} + 1)^2\cdot(1-z)^2}`

This can be simplified:

`\frac{4}{(\frac{z+1}{1-z} + \frac{1-z}{1-z})^2\cdot(1-z)^2}=\frac{4}{(\frac{2}{1-z})^2\cdot(1-z)^2}=\frac{4}{4}=1`