Differentiate y = (sqrt(1+x2)−1x) with respect to x.

Let $x = \tan\theta$.
Then $\sqrt{1+x^2} = \sqrt{1+\tan^2\theta} = \sqrt{\sec^2\theta} = |\sec\theta|$.
For simplicity, assume $x > 0$, which implies $\theta \in (0, \pi/2)$, so $\sec\theta > 0$.
Substitute these into the expression inside the $\arctan$:

$$\frac{\sqrt{1+x^2}-1}{x} = \frac{\sec\theta - 1}{\tan\theta}$$

Rewrite $\sec\theta$ and $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$:

$$\frac{\sec\theta - 1}{\tan\theta} = \frac{\frac{1}{\cos\theta} - 1}{\frac{\sin\theta}{\cos\theta}} = \frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1-\cos\theta}{\sin\theta}$$

Now, use the half-angle identities: $1-\cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$ and $\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$.

$$\frac{1-\cos\theta}{\sin\theta} = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right)$$

$$y = \arctan\left(\tan\left(\frac{\theta}{2}\right)\right)$$

Since $x = \tan\theta$, for $x > 0$, we have $\theta = \arctan x$, and $\theta \in (0, \pi/2)$. This means $\frac{\theta}{2} \in (0, \pi/4)$. In this interval, $\arctan(\tan u) = u$.
Therefore,

$$y = \frac{\theta}{2}$$

Since $\theta = \arctan x$, substitute this back into the expression for $y$:

$$y = \frac{1}{2}\arctan x$$

This simplification holds for $x \neq 0$.

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}\arctan x\right)$$

Recall that $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$.

$$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2}$$ $$\frac{dy}{dx} = \frac{1}{2(1+x^2)}$$

The final answer is $\frac{1}{2(1+x^2)}$.