Continuity and Differentiability — Class 12 Maths Solution

Let $y = {\sec ^{ - 1}}\left( {\frac{1}{{4{x^3} - 3x}}} \right)$
On putting $x = \cos \theta$ in Eq. (i), we get
$y = {\sec ^{ - 1}}\frac{1}{{4{{\cos }^3}\theta - 3\cos \theta }}$ …….(i)

$= {\sec ^{ - 1}}\frac{1}{{\cos 3\theta }}$

$= {\sec ^{ - 1}}(\sec 3\theta ) = 3\theta$

$= 3{\cos ^{ - 1}}x$

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {3{{\cos }^{ - 1}}x} \right)$

$= 3 \cdot \frac{{ - 1}}{{\sqrt {1 - {x^2}} }}$

$\frac{dy}{dx} = -3 \cdot \frac{{ 1}}{{\sqrt {1 - {x^2}} }}$