Question

please help me wt this qs​

Answer 1

Given

$\sf\to y =(cot^{-1}x)^2$

To Prove

$\sf\to (x^2+1)^2\dfrac{d^2y}{dx^2} +2x(x^2+1)\dfrac{dy}{dx} = 2$

Now Take

$\sf\to y =(cot^{-1}x)^2$

$\sf\to\dfrac{dy}{dx} =(cot^{-1}x)^2$

By using the Chain rule we get

$\sf\to\dfrac{dy}{dx} = 2(cot^{-1}x)\times\dfrac{-1}{x^2+1} =\dfrac{-2(cot^{-1}x)}{x^2+1}$

Now we have to find

$\sf\to \dfrac{d^2y}{dx^2} =\dfrac{-2(cot^{-1}x)}{x^2+1}$

By using u/v method we get

$\sf\to \dfrac{d^2y}{dx^2} = \dfrac{(x^2+1)\times -2\dfrac{d(cot^{-1}x)}{dx}-(-2)(cot^{-1}x)\dfrac{d(x^2+1)}{dx} }{(x^2+1)^2}$

$\sf\to \dfrac{d^2y}{dx^2} =\dfrac{(x^2+1)\times\dfrac{-1\times-2}{(x^2+1)}+2cot^{-1}x\times2x }{(x^2+1)^2}$

$\sf\to \dfrac{d^2y}{dx^2} = \dfrac{2+4xcot^{-1}x}{(x^2+1)^2}$

Now put the value on

$\sf\to (x^2+1)^2\dfrac{d^2y}{dx^2} +2x(x^2+1)\dfrac{dy}{dx}$

We have

$\sf\to\dfrac{dy}{dx} =\dfrac{-2(cot^{-1}x)}{x^2+1}$

$\sf\to \dfrac{d^2y}{dx^2} = \dfrac{2+4xcot^{-1}x}{(x^2+1)^2}$

Put the value

$\sf\to (x^2+1)^2\times \dfrac{2+4xcot^{-1}x}{(x^2+1)^2} +2x(x^2+1)\times\dfrac{-2(cot^{-1}x)}{x^2+1}$

$\sf\to \cancel{(x^2+1)^2}\times \dfrac{2+4xcot^{-1}x}{\cancel{(x^2+1)^2}}+2x\cancel{(x^2+1)}\times\dfrac{-2(cot^{-1}x)}{\cancel{x^2+1}}$

$\sf\to 2+4cot^{-1}x-4xcot^{-1}x$

$\sf\to 2$

Hence Proved