class 12 maths continuity and differentiability

$y = {\tan ^{ - 1}}\left( {\cfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), - \cfrac{1}{{\sqrt 3 }} < x < \cfrac{1}{{\sqrt 3 }}$

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📘 Continuity and Differentiability NCERT Ex.5.3 ,Q.10,Page 169 SA

$y = {\tan ^{ - 1}}\left( {\cfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), - \cfrac{1}{{\sqrt 3 }} < x < \cfrac{1}{{\sqrt 3 }}$

Official Solution

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$y = {\tan ^{ - 1}}\left( {\cfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right)$
Putting $x = \tan \theta ,$ we get

$y = {\tan ^{ - 1}}\left( {\cfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)$

$\Rightarrow$ $y = {\tan ^{ - 1}}(\tan 3\theta )$
$\Rightarrow$ $y = 3\theta \Rightarrow y = 3{\tan ^{ - 1}}x$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{3}{{1 + {x^2}}}$

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