Question

y=x+1/x-1 find d^2y/dx^2​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=\dfrac{x+1}{x-1}}$

$\tt{\implies\,y(x-1)=x+1}$

$\tt{\implies\,xy-y=x+1}$

$\tt{\implies\,y+x\,\dfrac{dy}{dx}-\dfrac{dy}{dx}=1}$

$\tt{\implies\,y+(x-1)\dfrac{dy}{dx}=1}$

$\tt{\implies\,\dfrac{dy}{dx}=\dfrac{1-y}{x-1}\,\,\,\,...(1)}$

Now,

$\tt{\implies\,\dfrac{dy}{dx}+\dfrac{dy}{dx}+(x-1)\dfrac{d^2y}{dx^2}=0}$

$\tt{\implies\,2\dfrac{dy}{dx}+(x-1)\dfrac{d^2y}{dx^2}=0}$

$\tt{\implies\,(x-1)\dfrac{d^2y}{dx^2}=-2\dfrac{dy}{dx}}$

$\tt{\implies\,(x-1)\dfrac{d^2y}{dx^2}=-2\cdot\dfrac{1-y}{x-1}}$

$\tt{\implies\,\dfrac{d^2y}{dx^2}=2\cdot\dfrac{y-1}{\left(x-1\right)^2}}$

$\tt{\implies\,\dfrac{d^2y}{dx^2}=2\cdot\dfrac{\dfrac{x+1}{x-1}-1}{\left(x-1\right)^2}}$

$\tt{\implies\,\dfrac{d^2y}{dx^2}=2\cdot\dfrac{\dfrac{x+1-(x-1)}{x-1}}{\left(x-1\right)^2}}$

$\tt{\implies\,\dfrac{d^2y}{dx^2}=2\cdot\dfrac{x+1-x+1}{\left(x-1\right)^3}}$

$\tt{\implies\,\dfrac{d^2y}{dx^2}=2\cdot\dfrac{1+1}{\left(x-1\right)^3}}$

$\tt{\implies\,\dfrac{d^2y}{dx^2}=2\cdot\dfrac{2}{\left(x-1\right)^3}}$

$\tt{\implies\,\dfrac{d^2y}{dx^2}=\dfrac{4}{\left(x-1\right)^3}}$

Answer 2

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

Given function is

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

can be rewritten as

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

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

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

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

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

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

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

We know,

$\boxed{\tt{ \dfrac{d}{dx} k = 0}}$

and

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1} }}$

So, using this, we get

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

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

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

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

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

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

$\rm :\longmapsto\: \dfrac{ {d}^{2} y}{d {x}^{2} } = 4{(x - 1)}^{ - 3}(1 - 0)$

$\rm :\longmapsto\: \boxed{\tt{ \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{4}{ {(x - 1)}^{3} } }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

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