Question
If $\text{y}=(\cot^{-1}\text{x})^2$ prove that $\text{y}^2(\text{x}^2+1)^2+2\text{x}(\text{x}^2+1)\text{y}_1=2.$
Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}_1=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}$
$=\frac{-2\cot^{-1}\text{x}}{1+\text{x}^2}\ (\text{chain rule})$
$\Rightarrow(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=-2\cot^{-1}\text{x}$
Differentiating w.r.t.x,
$\Rightarrow(1+\text{x}^2)\text{y}^2+2\text{xy}_1=+2\Big(\frac{+1}{1+\text{x}^2}\Big)$
(Multiplication rule on LHS)
$\Rightarrow(1+\text{x}^2)^2\text{y}_2+2\text{x}(1+\text{x}^2)\text{y}_1=2$
Hence proved
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$\frac{\text{dy}}{\text{dx}}-\text{y}=\cos2\text{x}$