Differentiate the following w.r.t. x:


Differentiate the following w.r.t. x:

$\sin \left(\tan ^{-1} e^{-x}\right)$


Let $y=\sin \left(\tan ^{-1} e^{-x}\right)$

By using the chain rule, we obtain

$\frac{d y}{d x}=\frac{d}{d x}\left[\sin \left(\tan ^{-1} e^{-x}\right)\right]$

$=\cos \left(\tan ^{-1} e^{-x}\right) \cdot \frac{d}{d x}\left(\tan ^{-1} e^{-x}\right)$

$=\cos \left(\tan ^{-1} e^{-x}\right) \cdot \frac{1}{1+\left(e^{-x}\right)^{2}} \cdot \frac{d}{d x}\left(e^{-x}\right)$

$=\frac{\cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \cdot e^{-x} \cdot \frac{d}{d x}(-x)$

$=\frac{e^{-x} \cos \left(\tan ^{-1} e^{-x}\right)}{1+e^{-2 x}} \times(-1)$

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

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