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

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(A) Given: $y = 	an^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} ight)$.Substitute $x = 	an 	heta$, which implies $	heta = 	an^{-1} x$.Then, $y = 	an^

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$1/4$

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(A) Given: $y = 	an^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} ight)$.Substitute $x = 	an 	heta$, which implies $	heta = 	an^{-1} x$.Then, $y = 	an^{-1} \left( \frac{\sqrt{1+	an^2 	heta}-1}{	an 	heta} ight) = 	an^{-1} \left( \frac{\sec 	heta - 1}{	an 	heta} ight)$.Using trigonometric identities, $\frac{\sec 	heta - 1}{	an 	heta} = \frac{1-\cos 	heta}{\sin 	heta} = \frac{2 \sin^2 (	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)} = 	an(	heta/2)$.So, $y = 	an^{-1} (	an(	heta/2)) = 	heta/2 = \frac{1}{2} 	an^{-1} x$.Differentiating with respect to $x$, we get $y' = \frac{1}{2(1+x^2)}$.At $x = 1$, $y'(1) = \frac{1}{2(1+1^2)} = \frac{1}{2(2)} = \frac{1}{4}$.

If $y = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$, then find the value of $y'(1)$.

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