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{\

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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 $x$:$y' = \frac{1}{2} \cdot \frac{1}{1+(ax)^2} \cdot a = \frac{a}{2(1+a^2x^2)}$.Thus, $2(1+a^2x^2)y' = a$.Differentiating both sides with respect to $x$ using the product rule:$2[(1+a^2x^2)y'' + y'(2a^2x)] = 0$.$(1+a^2x^2)y'' + 2a^2xy' = 0$.

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

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