If y = (tan^{-1} x)^2 ,then (x^2 + 1)^2 y_2 | vedclass

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(D) Given $ y = (	an^{-1} x)^2 $.First,differentiate with respect to $ x $:$ y_1 = \frac{dy}{dx} = 2(	an^{-1} x) \cdot \frac{1}{1+x^2} $.This implie

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$ 2 $

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(D) Given $ y = (	an^{-1} x)^2 $.First,differentiate with respect to $ x $:$ y_1 = \frac{dy}{dx} = 2(	an^{-1} x) \cdot \frac{1}{1+x^2} $.This implies $ (1+x^2) y_1 = 2 	an^{-1} x $.Differentiate both sides again with respect to $ x $ using the product rule:$ (1+x^2) y_2 + y_1(2x) = 2 \cdot \frac{1}{1+x^2} $.Multiply the entire equation by $ (1+x^2) $:$ (1+x^2)^2 y_2 + 2x(1+x^2) y_1 = 2 $.Thus,the value is $ 2 $.

If $ y = (\tan^{-1} x)^2 $,then $ (x^2 + 1)^2 y_2 + 2x(x^2 + 1) y_1 $ is equal to

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