Continuity and Differentiability — Class 12 Maths Solution

Let $u = {\cos ^{ - 1}}\left( {2{x^2} - 1} \right)$ and $v = {\cos ^{ - 1}}x$

therefore,$\frac{{dv}}{{dx}} = \frac{{ + - 1}}{{\sqrt {1 - {{\left( {2{x^2} - 1} \right)}^2}} }} \cdot 4x = \frac{{ - 4x}}{{\sqrt {1 - \left( {4{x^4} + 1 - 4{x^2}} \right)} }}$

$= \frac{{ - 4x}}{{\sqrt { - 4{x^4} + 4{x^2}} }} = \frac{{ - 4x}}{{\sqrt {4{x^2}\left( {1 - {x^2}} \right)} }}$
$= \frac{{ - 2}}{{\sqrt {1 - {x^2}} }}$

and $\frac{{du}}{{dx}} = \frac{{ - 1}}{{\sqrt {1 - {x^2}} }}$
therefore,$\frac{{dx}}{{dv}} = \frac{{du/dx}}{{dv/dx}} = \frac{{ - 2/\sqrt {1 - {x^2}} }}{{ - 1/\sqrt {1 - {x^2}} }} = 2$