Differentiate the following with respect to x | vedclass

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Let $y = \sin \left(	an ^{-1} e^{-x}ight)$.By using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left[ \sin \

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

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

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