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Parametric Differentiation

If x=tanh-1...

Question

If $x = {tanh}^{- 1} y$ , then $(\frac{1 + y}{1 - y}) =$

A

e^{x}

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B

e^{4 x}

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C

e^{2 x}

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D

e^{3 x}

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Solution

The correct option is C $e^{2 x}$
$x = tan h^{- 1} y$
$tan x = y = \frac{e^{+ x} - e^{- x}}{e^{x} + e^{- x}} [\frac{s i n h y}{cos h y} = \frac{e^{+ x} - e^{- x}}{e^{x} + e^{- x}}]$
$∴ 1 + y = \frac{2 e^{x}}{e^{x} + e^{- x}}$
$1 - y = \frac{2 e^{- x}}{e^{x} + e^{- x}}$
$∴ \frac{1 + y}{1 - y} = e^{2 x}$
$l n (\frac{1 + y}{1 - y}) = l n e^{2 x} = 2 x$

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