The problem seems to ask to find the derivative of the function $y = f(x) = \ln(x + \sqrt{1+x^2})$.

1. Problem Description

The problem seems to ask to find the derivative of the function $y = f(x) = \ln(x + \sqrt{1+x^2})$.

2. Solution Steps

To find the derivative of $y = \ln(x + \sqrt{1+x^2})$, we can use the chain rule.

Let $u = x + \sqrt{1+x^2}$. Then $y = \ln(u)$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = \frac{1}{u} = \frac{1}{x + \sqrt{1+x^2}}$

Now, we need to find $\frac{du}{dx}$.

$u = x + \sqrt{1+x^2}$

$\frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{1+x^2})$

$\frac{d}{dx}(x) = 1$

Let $v = 1+x^2$. Then $\sqrt{1+x^2} = \sqrt{v} = v^{1/2}$.

$\frac{d}{dx}(\sqrt{1+x^2}) = \frac{d}{dv}(v^{1/2}) \cdot \frac{dv}{dx}$

$\frac{d}{dv}(v^{1/2}) = \frac{1}{2}v^{-1/2} = \frac{1}{2\sqrt{v}} = \frac{1}{2\sqrt{1+x^2}}$

$\frac{dv}{dx} = \frac{d}{dx}(1+x^2) = 2x$

So, $\frac{d}{dx}(\sqrt{1+x^2}) = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$

Therefore, $\frac{du}{dx} = 1 + \frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}$

Now, we can find $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{x + \sqrt{1+x^2}} \cdot \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}$

3. Final Answer

$\frac{dy}{dx} = \frac{1}{\sqrt{1+x^2}}$