Continuity and Differentiability — Class 12 Maths Solution

ncert exercise SA NCERT Ex.5.4 ,Q.2,Page 174
Question

${e^{{{\sin }^{ - 1}}x}}$

Step-by-step Solution

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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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Continuity and Differentiability. Curated by Sachin Sharma. Free for all students.