Find $\frac{dy}{dx}$ where $y=\sin(\tan^{-1}e^x)$1$\frac{e^2x}{1+e^{2x}}\cos(\tan^{-1}e^x)$2$\frac{e^x}{1+e^{2x}}\cos(\tan^{-1}e^x)$3$\frac{e^x}{1+e^{2x}}\sin(\tan^{-1}e^x)$4$\frac{e^x}{1+e^{x}}\cos(\tan^{-1}e^x)$Answer & explanation+Correct answer: option 2$\frac{dy}{dx}=\cos(tan^{-1}e^x)\frac{d}{dx}(\tan^{-1}e^x)=\frac{e^x}{1+e^{2x}}\cos(\tan^{-1}e^x)$
