Continuity and Differentiability — Class 12 Maths Solution

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

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}(\sin ({\tan ^{ - 1}}({e^{ - x}}))) = \cos ({\tan ^{ - 1}}{e^{ - x}})\cfrac{d}{{dx}}({\tan ^{ - 1}}({e^{ - x}}))$

$= \cos ({\tan ^{ - 1}}{e^{ - x}}) \cdot \cfrac{1}{{(1 + {e^{ - 2x}})}} \cdot \cfrac{d}{{dx}}{e^{ - x}}$

$= \cos ({\tan ^{ - 1}}{e^{ - x}}) \cdot \cfrac{{ - {e^{ - x}}}}{{(1 + {e^{ - 2x}})}} = \cfrac{{ - {e^{ - x}}\cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {e^{ - 2x}}}}$