If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, then
Solution
<p>Let $${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$$</p>
<p>$\therefore$ $y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$</p>
<p>$= {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$</p>
<p>$\therefore$ $y = {\pi \over 4} - {\theta \over 2}$</p>
<p>$y = {\pi \over 4} - {{{x^3}} \over 2}$</p>
<p>$\therefore$ $y' = {{ - 3{x^2}} \over 2}$</p>
<p>$y'' = - 3x$</p>
<p>$\therefore$ ${x^2}y'' - 6y + {{3\pi } \over 2} = 0$</p>
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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