Differentiate the following with respect to $x$: $\sin \left(\tan ^{-1} e^{-x}\right)$

Let $y = \sin \left(\tan ^{-1} e^{-x}\right)$.
By using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left[ \sin \left( \tan^{-1} e^{-x} \right) \right]$
$= \cos \left( \tan^{-1} e^{-x} \right) \cdot \frac{d}{dx} \left( \tan^{-1} e^{-x} \right)$
$= \cos \left( \tan^{-1} e^{-x} \right) \cdot \frac{1}{1 + (e^{-x})^2} \cdot \frac{d}{dx} (e^{-x})$
$= \cos \left( \tan^{-1} e^{-x} \right) \cdot \frac{1}{1 + e^{-2x}} \cdot (e^{-x} \cdot -1)$
$= \frac{-e^{-x} \cos \left( \tan^{-1} e^{-x} \right)}{1 + e^{-2x}}$