Question #2c4f7

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Question #2c4f7

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Answer 1 · 2017-01-27T08:15:25

ans. the derivative is $\frac{- e^{- 2 y}}{{(x - 1)}^{2}}$ $(-e^(-2y))/(x-1)^2$ , where $y = ln \sqrt{\frac{x + 1}{x - 1}}$ $y =ln sqrt((x+1)/(x-1))$ .

Explanation:

see, at first let $y = ln \sqrt{\frac{x + 1}{x - 1}}$ $y =ln sqrt((x+1)/(x-1))$
or $e^{y} = \sqrt{\frac{x + 1}{x - 1}}$ $e^y = sqrt((x+1)/(x-1))$
or $e^{2 y}$ $e^(2y)$ = $\frac{x + 1}{x - 1}$ $(x+1)/(x-1)$ (sqaring)
now differenting the above w.r.t x we get ,
$\frac{d y}{d x} 2 e^{2 Y}$ $dy/dx 2e^(2Y)$ = $\frac{(x - 1) - (x + 1)}{{(x - 1)}^{2}}$ $((x-1)-(x+1))/(x-1)^2$ by $(d \frac{\frac{u}{v}}{d x}) m e t h o d$ $(d(u/v)/dx) method$
or $\frac{d y}{d x} e^{2 y}$ $dy/dx e^(2y)$ = $- \frac{1}{{(x - 1)}^{2}}$ $-1/(x-1)^2$
or we get $\frac{d y}{d x} = - \frac{e^{- 2 y}}{{(x - 1)}^{2}}$ $dy/dx =-e^(-2y)/(x-1)^2$ where $y = ln \sqrt{\frac{x + 1}{x - 1}}$ $y=lnsqrt((x+1)/(x-1))$ .

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Question #2c4f7

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