The derivative of $\sin(\tan^{-1}e^{2x})$ with respect to $x$ is:

$\frac{2e^{2x}\sin(\tan^{-1}e^{2x})}{1+e^{4x}}$ $\frac{2e^{2x}\cos(\tan^{-1}e^{2x})}{1+e^{4x}}$ $\frac{2e^{2x}\sin(\tan^{-1}e^{2x})}{1+e^{x^2}}$ $\frac{2e^{2x}\cos(\tan^{-1}e^{2x})}{1+e^{2x}}$

$\frac{2e^{2x}\cos(\tan^{-1}e^{2x})}{1+e^{4x}}$

$y=\sin(\tan^{-1}e^{2x})$ differentiating w.r.t x $\frac{dy}{dx}=\cos(\tan^{-1}e^{2x})\frac{d}{dx}(\tan^{-1}e^{2x})⇒\frac{dy}{dx}=\frac{\cos(\tan^{-1}e^{2x})}{1+e^{4x}}\frac{de^{2x}}{x}$ $=\frac{2e^{2x}\cos(\tan^{-1}e^{2x})}{1+e^{4x}}$