If $y = {({\tan ^{ - 1}}x)^2},$ show that ${({x^2} + 1)^2}{y_2} + 2x({x^2} + 1){y_1} = 2.$

Let $y = {({\tan ^{ - 1}}x)^2}$ ...(i)
Differentiating (i) w.r.t. x, we get

$\cfrac{{dy}}{{dx}} = 2{\tan ^{ - 1}}x \cdot \cfrac{1}{{(1 + {x^2})}}$ ….(ii)

Differentiating (ii) w.r.t.x, we get

$\cfrac{{{d^2}y}}{{d{x^2}}} = 2\left[ {{{\tan }^{ - 1}}x\cfrac{{(\{ 1 + {x^2}\} \cdot 0 - 2x)}}{{{{(1 + {x^2})}^2}}} + \cfrac{1}{{{{(1 + x)}^2}}} \cdot \cfrac{1}{{(1 + {x^2})}}} \right]$

$= 2\left[ {\cfrac{{ - 2x{{\tan }^{ - 1}}x}}{{{{(1 + {x^2})}^2}}} + \cfrac{1}{{{{(1 + {x^2})}^2}}}} \right] = 2\left[ {\cfrac{{ - 2x{{\tan }^{ - 1}}x + 1}}{{{{(1 + {x^2})}^2}}}} \right]$

Now, ${({x^2} + 1)^2}\cfrac{{{d^2}y}}{{d{x^2}}} + 2x({x^2} + 1)\cfrac{{dy}}{{dx}}$

$= {({x^2} + 1)^2} \cdot 2\left[ {\cfrac{{ - 2x{{\tan }^{ - 1}}x + 1}}{{{{(1 + {x^2})}^2}}}} \right] + 2x({x^2} + 1) \cdot 2{\tan ^{ - 1}}x\cfrac{1}{{(1 + {x^2})}}$

$= - 4x{\tan ^{ - 1}}x + 2 + 4x{\tan ^{ - 1}}x = 2$

Hence proved.