Let
\[
y = \sin\left(\tan^{-1} e^{2x}\right)
\]
Step 1: Use the chain rule for differentiation:
\[
\frac{dy}{dx} = \cos\left(\tan^{-1} e^{2x}\right) \cdot \frac{d}{dx} \left(\tan^{-1} e^{2x}\right)
\]
Step 2: Differentiate \(\tan^{-1} e^{2x}\) using the chain rule:
\[
\frac{d}{dx} \tan^{-1} u = \frac{1}{1+u^2} \frac{du}{dx}
\]
where \(u = e^{2x}\), so
\[
\frac{du}{dx} = 2e^{2x}
\]
Therefore,
\[
\frac{d}{dx} \left(\tan^{-1} e^{2x}\right) = \frac{2e^{2x}}{1 + \left(e^{2x}\right)^2} = \frac{2e^{2x}}{1 + e^{4x}}
\]
Step 3: Combine the results:
\[
\frac{dy}{dx} = \cos\left(\tan^{-1} e^{2x}\right) \cdot \frac{2e^{2x}}{1 + e^{4x}} = \frac{2e^{2x} \cos\left(\tan^{-1} e^{2x}\right)}{1 + e^{4x}}
\]