Let\[y = \sin\left(\tan^{-1} e^{2x}\right)\]To differentiate y with respect to x, we apply the chain rule:\[\frac{dy}{dx} = \cos\left(\tan^{-1} e^{2x}\right) \cdot \frac{d}{dx} \left(\tan^{-1} e^{2x}\right)\]Next, we differentiate \(\tan^{-1} e^{2x}\) using the chain rule. The derivative of \(\tan^{-1} u\) with respect to \(u\) is \(\frac{1}{1+u^2}\). Let \(u = e^{2x}\). Then \(\frac{du}{dx} = 2e^{2x}\). Applying the chain rule:\[\frac{d}{dx} \tan^{-1} u = \frac{1}{1+u^2} \frac{du}{dx}\]Substituting \(u = e^{2x}\) and \(\frac{du}{dx} = 2e^{2x}\):\[\frac{d}{dx} \left(\tan^{-1} e^{2x}\right) = \frac{1}{1+\left(e^{2x}\right)^2} \cdot 2e^{2x} = \frac{2e^{2x}}{1 + e^{4x}}\]Finally, we combine the results to find \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \cos\left(\tan^{-1} e^{2x}\right) \cdot \frac{2e^{2x}}{1 + e^{4x}} = \frac{2e^{2x} \cos\left(\tan^{-1} e^{2x}\right)}{1 + e^{4x}}\]