$g (x) = \frac{x - 1}{x + 1}, x = \frac{z + 1}{1 - z}$ $g(x)=(x-1)/(x+1), x=(z+1)/(1-z)$ .Then $\frac{d g}{d z} =$ $(dg)/dz=$ ?(Express it in z)

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$g (x) = \frac{x - 1}{x + 1}, x = \frac{z + 1}{1 - z}$ $g(x)=(x-1)/(x+1), x=(z+1)/(1-z)$ .Then $\frac{d g}{d z} =$ $(dg)/dz=$ ?(Express it in z)

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Answer 1 · 2018-06-29T04:34:30

$\frac{d g}{d z} = 1$ $\frac{dg}{dz}=1$

Explanation:

$\frac{d g}{d z} = \frac{d g}{d x} \cdot \frac{d x}{d z}$ $\frac{dg}{dz} = \frac{dg}{dx}\cdot\frac{dx}{dz}$

By quotient rule:

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

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

Therefore,

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

This can be simplified:

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

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$g (x) = \frac{x - 1}{x + 1}, x = \frac{z + 1}{1 - z}$ $g(x)=(x-1)/(x+1), x=(z+1)/(1-z)$ .Then $\frac{d g}{d z} =$ $(dg)/dz=$ ?(Express it in z)