If $$y = e^{nx}$$, then $$\frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2}$$ is equal to:

We are given that $$ y = e^{nx} $$. We need to find $$ \frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} $$.

First, we find $$ \frac{d^2y}{dx^2} $$. Since $$ y = e^{nx} $$, we start by finding the first derivative $$ \frac{dy}{dx} $$. The derivative of $$ e^{nx} $$ with respect to $$ x $$ is $$ n e^{nx} $$, so:

$$ \frac{dy}{dx} = n e^{nx} $$

But since $$ y = e^{nx} $$, we can write $$ \frac{dy}{dx} = n y $$. Now, to find the second derivative $$ \frac{d^2y}{dx^2} $$, we differentiate $$ \frac{dy}{dx} $$ again with respect to $$ x $$:

$$ \frac{d^2y}{dx^2} = \frac{d}{dx} (n y) $$

Using the chain rule, since $$ y $$ is a function of $$ x $$, we get:

$$ \frac{d}{dx} (n y) = n \frac{dy}{dx} = n \cdot (n y) = n^2 y $$

Alternatively, differentiating directly:

$$ \frac{d}{dx} (n e^{nx}) = n \cdot n e^{nx} = n^2 e^{nx} = n^2 y $$

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

Next, we need $$ \frac{d^2x}{dy^2} $$. This requires expressing $$ x $$ in terms of $$ y $$. Given $$ y = e^{nx} $$, we solve for $$ x $$:

Take the natural logarithm of both sides:

$$ \ln y = \ln(e^{nx}) $$

$$ \ln y = nx $$

Therefore,

$$ x = \frac{1}{n} \ln y $$

Now, find the first derivative $$ \frac{dx}{dy} $$:

$$ \frac{dx}{dy} = \frac{d}{dy} \left( \frac{1}{n} \ln y \right) = \frac{1}{n} \cdot \frac{1}{y} = \frac{1}{n y} $$

Now, find the second derivative $$ \frac{d^2x}{dy^2} $$ by differentiating $$ \frac{dx}{dy} $$ with respect to $$ y $$:

$$ \frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{1}{n y} \right) = \frac{1}{n} \cdot \frac{d}{dy} (y^{-1}) $$

The derivative of $$ y^{-1} $$ is $$ -y^{-2} $$, so:

$$ \frac{d^2x}{dy^2} = \frac{1}{n} \cdot (-y^{-2}) = -\frac{1}{n y^2} $$

Now, we multiply the two second derivatives:

$$ \frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = (n^2 y) \cdot \left( -\frac{1}{n y^2} \right) $$

Simplify the expression:

$$ = n^2 y \cdot \left( -\frac{1}{n y^2} \right) = -\frac{n^2}{n} \cdot \frac{y}{y^2} = -n \cdot \frac{1}{y} = -\frac{n}{y} $$

Since $$ y = e^{nx} $$, substitute back:

$$ -\frac{n}{y} = -\frac{n}{e^{nx}} = -n e^{-nx} $$

because $$ \frac{1}{e^{nx}} = e^{-nx} $$.

Therefore, $$ \frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = -n e^{-nx} $$.

Comparing with the options:

A. $$ ne^{-nx} $$

B. $$ -ne^{-nx} $$

C. $$ ne^{nx} $$

D. 1

Hence, the correct answer is Option B.