If y=tan ^{-1}left(frac{sqrt{1+a^2 x^2}-1}{a | vedclass

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(D) Given,$y=	an ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}ight)$.Put $ax = 	an 	heta$,then $	heta = 	an^{-1}(ax)$.$y = 	an^{-1}\left(\frac{\se

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(D) Given,$y=	an ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}ight)$.Put $ax = 	an 	heta$,then $	heta = 	an^{-1}(ax)$.$y = 	an^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ight) = 	an^{-1}\left(\frac{1-\cos 	heta}{\sin 	heta}ight) = 	an^{-1}\left(	an \frac{	heta}{2}ight) = \frac{	heta}{2} = \frac{1}{2} 	an^{-1}(ax)$.Differentiating with respect to $x$:$y^{\prime} = \frac{1}{2} \cdot \frac{1}{1+(ax)^2} \cdot a = \frac{a}{2(1+a^2x^2)}$.Thus,$2(1+a^2x^2)y^{\prime} = a$.Differentiating both sides with respect to $x$ using the product rule:$2 \left[ (1+a^2x^2)y^{\prime \prime} + y^{\prime} (2a^2x) ight] = 0$.Dividing by $2$:$(1+a^2x^2)y^{\prime \prime} + 2a^2x y^{\prime} = 0$.

If $y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$,then $\left(1+a^2 x^2\right) y^{\prime \prime}+2 a^2 x y^{\prime}$ is equal to

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