Continuity and Differentiability — Class 12 Maths Solution

ncert exercise SA NCERT Ex.5.3 ,Q.14,Page 169
Question

$y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), - \cfrac{1}{{\sqrt 2 }} < x < \cfrac{1}{{\sqrt 2 }}.$

Step-by-step Solution

Putting $x = \sin \theta ,$ we get

$y = {\sin ^{ - 1}}[2\sin \theta \sqrt {1 - {{\sin }^2}\theta } ] \Rightarrow y = {\sin ^{ - 1}}(\sin 2\theta ) \Rightarrow y = 2\theta$

$\Rightarrow$ $y = 2{\sin ^{ - 1}}x \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{2}{{\sqrt {1 - {x^2}} }}$

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