Question

find dy/dx if √1+x/1-x​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = \sqrt{\dfrac{1 + x}{1 - x} }$

On squaring both sides, we get

$\rm :\longmapsto\: {y}^{2} = \dfrac{1 + x}{1 - x}$

On taking log both sides, we get

$\rm :\longmapsto\: log \: {y}^{2} =log \: \bigg[ \dfrac{1 + x}{1 - x} \bigg]$

We know

$\boxed{ \tt{ \: log {x}^{y} = y \: logx}} \\ \\ and \\ \\ \boxed{ \tt{ \: log \frac{x}{y} = logx - logy}}$

So, using these, we get

$\rm :\longmapsto\:2 \: logy = log(1 + x) - log(1 - x)$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} (2 \: logy) = \dfrac{d}{dx} [log(1 + x) - log(1 - x)]$

$\rm :\longmapsto\:2\dfrac{d}{dx} (\: logy) = \dfrac{d}{dx}log(1 + x) - \dfrac{d}{dx} log(1 - x)$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} logx = \frac{1}{x} \: }}$

So, using this, we get

$\rm :\longmapsto\:2 \: \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{1 + x}\dfrac{d}{dx} (1 + x) - \dfrac{1}{1 - x}\dfrac{d}{dx} (1 - x)$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{1}{1 + x}(0 + 1) - \dfrac{1}{1 - x} (0 - 1)$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{1}{1 + x} + \dfrac{1}{1 - x}$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{1 - x + 1 + x}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{2}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\: \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{y}{(1 + x)(1 - x)}$

Or

$\rm :\longmapsto\:\dfrac{dy}{dx} = \sqrt{\dfrac{1 + x}{1 - x} } \times \dfrac{1}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 - x} \: \sqrt{1 + x} \: (1 - x)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 + x} \: \: {\bigg(1 - x \bigg) }^{\dfrac{3}{2} } }$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = \sqrt{\dfrac{1 + x}{1 - x} }$

On squaring both sides, we get

$\rm :\longmapsto\: {y}^{2} = \dfrac{1 + x}{1 - x}$

On taking log both sides, we get

$\rm :\longmapsto\: log \: {y}^{2} =log \: \bigg[ \dfrac{1 + x}{1 - x} \bigg]$

We know

$\boxed{ \tt{ \: log {x}^{y} = y \: logx}} \\ \\ and \\ \\ \boxed{ \tt{ \: log \frac{x}{y} = logx - logy}}$

So, using these, we get

$\rm :\longmapsto\:2 \: logy = log(1 + x) - log(1 - x)$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} (2 \: logy) = \dfrac{d}{dx} [log(1 + x) - log(1 - x)]$

$\rm :\longmapsto\:2\dfrac{d}{dx} (\: logy) = \dfrac{d}{dx}log(1 + x) - \dfrac{d}{dx} log(1 - x)$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} logx = \frac{1}{x} \: }}$

So, using this, we get

$\rm :\longmapsto\:2 \: \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{1 + x}\dfrac{d}{dx} (1 + x) - \dfrac{1}{1 - x}\dfrac{d}{dx} (1 - x)$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{1}{1 + x}(0 + 1) - \dfrac{1}{1 - x} (0 - 1)$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{1}{1 + x} + \dfrac{1}{1 - x}$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{1 - x + 1 + x}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\: \dfrac{2}{y}\dfrac{dy}{dx} = \dfrac{2}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\: \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{y}{(1 + x)(1 - x)}$

Or

$\rm :\longmapsto\:\dfrac{dy}{dx} = \sqrt{\dfrac{1 + x}{1 - x} } \times \dfrac{1}{(1 + x)(1 - x)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 - x} \: \sqrt{1 + x} \: (1 - x)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 + x} \: \: {\bigg(1 - x \bigg) }^{\dfrac{3}{2} } }$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$