To solve this problem, we are tasked with finding $\frac{d^2y}{dx^2}$, given the parametric equations:

$x = \arccos(e^t)$ $y = \arcsin(e^{2t}).$

Step 1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

The derivatives of the given functions with respect to $t$ are:

1. For $x = \arccos(e^t)$: $\frac{dx}{dt} = \frac{-1}{\sqrt{1 - (e^t)^2}} \cdot e^t = \frac{-e^t}{\sqrt{1 - e^{2t}}}.$

2. For $y = \arcsin(e^{2t})$: $\frac{dy}{dt} = \frac{1}{\sqrt{1 - (e^{2t})^2}} \cdot 2e^{2t} = \frac{2e^{2t}}{\sqrt{1 - e^{4t}}}.$

Step 2: Find $\frac{dy}{dx}$

Using the chain rule, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, we substitute:

$\frac{dy}{dx} = \frac{\frac{2e^{2t}}{\sqrt{1 - e^{4t}}}}{\frac{-e^t}{\sqrt{1 - e^{2t}}}}.$

Simplify: $\frac{dy}{dx} = \frac{2e^{2t} \cdot \sqrt{1 - e^{2t}}}{-e^t \cdot \sqrt{1 - e^{4t}}}.$

Combine terms: $\frac{dy}{dx} = \frac{-2e^{2t} \sqrt{1 - e^{2t}}}{e^t \sqrt{(1 - e^{2t})(1 + e^{2t})}}.$

Simplify further: $\frac{dy}{dx} = \frac{-2e^{2t}}{e^t \sqrt{1 + e^{2t}}}.$

Finally: $\frac{dy}{dx} = \frac{-2e^t}{\sqrt{1 + e^{2t}}}.$

Step 3: Find $\frac{d^2y}{dx^2}$

To compute $\frac{d^2y}{dx^2}$, we use the formula: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.$

1. Differentiate $\frac{dy}{dx} = \frac{-2e^t}{\sqrt{1 + e^{2t}}}$ with respect to $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt} \left(-2e^t (1 + e^{2t})^{-1/2} \right).$

   Using the product rule and chain rule: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = -2e^t \cdot \frac{-1}{2}(1 + e^{2t})^{-3/2} \cdot 2e^{2t} + (-2)e^t.$

   Simplify: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2e^t \cdot e^{2t}}{(1 + e^{2t})^{3/2}} - 2e^t.$

2. Divide by $\frac{dx}{dt} = \frac{-e^t}{\sqrt{1 - e^{2t}}}$: $\frac{d^2y}{dx^2} = \frac{\frac{2e^t \cdot e^{2t}}{(1 + e^{2t})^{3/2}} - 2e^t}{\frac{-e^t}{\sqrt{1 - e^{2t}}}}.$

   Simplify: $\frac{d^2y}{dx^2} = \frac{\sqrt{1 - e^{2t}} \left(\frac{2e^t \cdot e^{2t}}{(1 + e^{2t})^{3/2}} - 2e^t\right)}{-e^t}.$

   Simplify further as needed for exact expressions.

Would you like to proceed with further simplifications or detailed explanations?

Related Questions:

1. What are some real-world applications of second derivatives in parametric equations?
2. How does the chain rule apply in finding higher-order derivatives?
3. What are the differences between implicit and parametric differentiation?
4. Can you demonstrate similar problems with trigonometric functions?
5. How can $\frac{d^2y}{dx^2}$ be interpreted geometrically?

Tip: Always check intermediate derivatives for errors when working with complex chain rules to ensure final expressions are correct.