If $$x = \int_0^y \frac{dt}{\sqrt{1+t^2}}$$, then $$\frac{d^2y}{dx^2}$$ is equal to :

We are given that $$ x = \int_0^y \frac{dt}{\sqrt{1+t^2}} $$. We need to find $$ \frac{d^2y}{dx^2} $$.

First, recall the Fundamental Theorem of Calculus. If $$ x = \int_a^y f(t) dt $$, then $$ \frac{dx}{dy} = f(y) $$. Here, the integrand is $$ f(t) = \frac{1}{\sqrt{1+t^2}} $$, and the lower limit is 0, which is a constant. So, differentiating both sides with respect to $$ y $$:

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

Now, we need $$ \frac{d^2y}{dx^2} $$, which is $$ \frac{d}{dx} \left( \frac{dy}{dx} \right) $$. First, find $$ \frac{dy}{dx} $$. Since $$ \frac{dy}{dx} $$ is the reciprocal of $$ \frac{dx}{dy} $$ (as long as $$ \frac{dx}{dy} \neq 0 $$):

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

So, $$ \frac{dy}{dx} = \sqrt{1+y^2} $$.

Next, find the second derivative by differentiating $$ \frac{dy}{dx} $$ with respect to $$ x $$:

$$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \sqrt{1+y^2} \right) $$

Since $$ y $$ is a function of $$ x $$, use the chain rule. Let $$ u = 1 + y^2 $$, so $$ \sqrt{1+y^2} = u^{1/2} $$. Then:

$$ \frac{d}{dx} \left( u^{1/2} \right) = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx} $$

Now, $$ \frac{du}{dx} = \frac{d}{dx}(1 + y^2) = 0 + 2y \frac{dy}{dx} $$, because the derivative of $$ y^2 $$ with respect to $$ x $$ is $$ 2y \frac{dy}{dx} $$. So:

$$ \frac{du}{dx} = 2y \frac{dy}{dx} $$

Substitute back:

$$ \frac{d}{dx} \left( \sqrt{1+y^2} \right) = \frac{1}{2} (1+y^2)^{-1/2} \cdot 2y \frac{dy}{dx} $$

Simplify the constants:

$$ = \frac{1}{2} \cdot 2y \cdot (1+y^2)^{-1/2} \cdot \frac{dy}{dx} = y (1+y^2)^{-1/2} \frac{dy}{dx} $$

We already know $$ \frac{dy}{dx} = \sqrt{1+y^2} = (1+y^2)^{1/2} $$, so substitute:

$$ = y (1+y^2)^{-1/2} \cdot (1+y^2)^{1/2} $$

Using the property of exponents $$ (1+y^2)^{-1/2} \cdot (1+y^2)^{1/2} = (1+y^2)^{0} = 1 $$:

$$ = y \cdot 1 = y $$

Therefore, $$ \frac{d^2y}{dx^2} = y $$.

Now, comparing with the options:

A. $$ y $$

B. $$ \sqrt{1 + y^2} $$

C. $$ \frac{x}{\sqrt{1+y^2}} $$

D. $$ y^2 $$

Hence, the correct answer is Option A.