Find the derivative of {sec ^{ - 1}}left( {fr | vedclass

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(D) Let $y = {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} ight) = {\cos ^{ - 1}}(2{x^2} - 1)$.Using the substitution $x = \cos 	heta$,we get $y =

Vedclass · Accepted Answer

$0$

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(D) Let $y = {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} ight) = {\cos ^{ - 1}}(2{x^2} - 1)$.Using the substitution $x = \cos 	heta$,we get $y = {\cos ^{ - 1}}(\cos 2	heta) = 2	heta = 2{\cos ^{ - 1}}x$.Now,let $z = \sqrt {1 + 3x} $.We need to find $\frac{{dy}}{{dz}} = \frac{{dy/dx}}{{dz/dx}}$.Calculating the derivatives:$\frac{{dy}}{{dx}} = 2 	imes \left( { - \frac{1}{{\sqrt {1 - {x^2}} }}} ight) = - \frac{2}{{\sqrt {1 - {x^2}} }}$.$\frac{{dz}}{{dx}} = \frac{1}{{2\sqrt {1 + 3x} }} 	imes 3 = \frac{3}{{2\sqrt {1 + 3x} }}$.At $x = - \frac{1}{3}$,the term $\sqrt {1 + 3x} = \sqrt {1 + 3(-1/3)} = \sqrt 0 = 0$.Since $\frac{{dz}}{{dx}}$ involves division by $\sqrt {1 + 3x}$,the derivative $\frac{{dz}}{{dx}}$ is undefined at $x = - \frac{1}{3}$.However,evaluating the limit or observing the function behavior,the derivative of the composition at the boundary point where the denominator of the second function becomes zero is typically considered $0$ in this specific context of multiple-choice questions where the derivative of the numerator is finite and the denominator approaches infinity.Thus,the correct option is $D$.

Find the derivative of ${\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right)$ with respect to $\sqrt {1 + 3x} $ at $x = - \frac{1}{3}$.

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