${e^{{{\sin }^{ - 1}}x}}$
${e^{{{\sin }^{ - 1}}x}}$
Official Solution
VVidaara Team
✓ Verified solution
NCERT & Exemplar
Let ${e^{{{\sin }^{ - 1}}x}}=y$
therefore, $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}({e^{{{\sin }^{ - 1}}x}}) = {e^{{{\sin }^{ - 1}}x}}\cfrac{d}{{dx}}({\sin ^{ - 1}}x)$
$= {e^{{{\sin }^{ - 1}}x}} \cdot \cfrac{1}{{\sqrt {1 - {x^2}} }},x \in ( - 1,1)$
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