Differentiate the given function w.r.t. $x$ .
$y = sin ({tan}^{- 1} e^{- x})$

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Solution

Let $sin ({tan}^{- 1} e^{- x})$
Thus by chain rule,
$\frac{d y}{d x} = \frac{d}{d x} [sin ({tan}^{- 1} e^{- x})]$
$= cos ({tan}^{- 1} e^{- x}) . \frac{d}{d x} ({tan}^{- 1} e^{- x})$
$= cos ({tan}^{- 1} e^{- x}) . \frac{1}{1 + (e^{- x})^{2}} \frac{d}{d x} (e^{- x})$
$= \frac{cos ({tan}^{- 1} e^{- x})}{1 + e^{- 2 x}} . e^{- x} . \frac{d}{d x} (- x)$
$= \frac{e^{- x} cos ({tan}^{- 1} e^{- x})}{1 + e^{- 2 x}} . (- 1)$
$= \frac{- e^{- x} cos ({tan}^{- 1} e^{- x})}{1 + e^{- 2 x}}$

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