Let's solve the problem step by step. The problem requires finding $\frac{dy}{dx}$, given:

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

Solution:

To find $\frac{dy}{dx}$, we use the chain rule: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$

Step 1: Differentiate $x = \arccos(e^t)$

The derivative of $\arccos(u)$ with respect to $t$ is: [ \frac{d}{dt}[\arccos(e^t)] = \frac{-1}{\sqrt{1 - (e^t)^2}} \cdot \frac{d}{dt}[e^t], ] where $\frac{d}{dt}[e^t] = e^t$. Thus: [ \frac{dx}{dt} = \frac{-e^t}{\sqrt{1 - e^{2