If y = (tan^{-1} 2x)^2 + (cot^{-1} 2x)^2,then | vedclass

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(B) Given $y = (	an^{-1} 2x)^2 + (\cot^{-1} 2x)^2$. Differentiating with respect to $x$: $y' = 2(	an^{-1} 2x) \cdot \frac{2}{1 + 4x^2} + 2(\cot^{-1}

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$-8x(1 + 4x^2) y'$

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(B) Given $y = (	an^{-1} 2x)^2 + (\cot^{-1} 2x)^2$. Differentiating with respect to $x$: $y' = 2(	an^{-1} 2x) \cdot \frac{2}{1 + 4x^2} + 2(\cot^{-1} 2x) \cdot \frac{-2}{1 + 4x^2}$ $y' = \frac{4(	an^{-1} 2x - \cot^{-1} 2x)}{1 + 4x^2}$ $(1 + 4x^2) y' = 4(	an^{-1} 2x - \cot^{-1} 2x)$ Differentiating again with respect to $x$: $8x y' + (1 + 4x^2) y'' = 4 \left( \frac{2}{1 + 4x^2} - \frac{-2}{1 + 4x^2} ight)$ $8x y' + (1 + 4x^2) y'' = 4 \left( \frac{4}{1 + 4x^2} ight) = \frac{16}{1 + 4x^2}$ Multiply both sides by $(1 + 4x^2)$: $8x(1 + 4x^2) y' + (1 + 4x^2)^2 y'' = 16$ $(1 + 4x^2)^2 y'' - 16 = -8x(1 + 4x^2) y'$

If $y = (\tan^{-1} 2x)^2 + (\cot^{-1} 2x)^2$,then $(1 + 4x^2)^2 y'' - 16 =$

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