Question:

Step 1: Simplify the given expression using a trigonometric substitution. Let $x = \tan \theta$. Then $\sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sec \theta$. Substituting into the original expression: $$y = 2\tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)$$ Step 2: Simplify the argument of the inverse tangent function. We need to simplify $\frac{\sec \theta - 1}{\tan \theta}$: $$\frac{\sec \theta - 1}{\tan \theta} = \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} = \frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{1-\cos \theta}{\sin \theta}$$ Using the half-angle identities $1 - \cos \theta = 2\sin^2(\theta/2)$ and $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$: $$\frac{1-\cos \theta}{\sin \theta} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan\left(\frac{\theta}{2}\right)$$ Step 3: Simplify the expression for y. Substituting back: $$y = 2\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = 2 \cdot \frac{\theta}{2} = \theta$$ Since $\theta = \tan^{-1}(x)$: $$y = \tan^{-1}(x)$$ Step 4: Find the first derivative. Differentiating with respect to $x$: $$\frac{dy}{dx} = \frac{1}{1+x^2}$$ Step 5: Find the second derivative. Differentiating again using the chain rule: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{1+x^2}\right) = \frac{d}{dx}(1+x^2)^{-1} = -1 \cdot (1+x^2)^{-2} \cdot 2x = \frac{-2x}{(1+x^2)^2}$$ Step 6: Evaluate the second derivative at $x = 2$. Substituting $x = 2$: $$\frac{d^2y}{dx^2}\bigg|_{x=2} = \frac{-2(2)}{(1+2^2)^2} = \frac{-4}{(1+4)^2} = \frac{-4}{25}$$ **Final Answer:** The value of $\frac{d^2y}{dx^2}$ at $x = 2$ is $\boxed{-\frac{4}{25}}$, which corresponds to **Option 2**.