The differential coefficient of {sec ^{ - 1}} | vedclass

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Question

(B) Let ${y_1} = {\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} ight)$ and ${y_2} = \sqrt {1 - {x^2}} $. Using the substitution $x = \cos 	heta$,we ha

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let ${y_1} = {\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} ight)$ and ${y_2} = \sqrt {1 - {x^2}} $. Using the substitution $x = \cos 	heta$,we have ${y_1} = {\sec ^{ - 1}}\left( \frac{1}{{2\cos^2 	heta - 1}} ight) = {\sec ^{ - 1}}(\sec 2	heta) = 2	heta = 2\cos^{-1}x$. Now,differentiate ${y_1}$ with respect to $x$: $\frac{{d{y_1}}}{{dx}} = 2 	imes \left( -\frac{1}{\sqrt{1-x^2}} ight) = -\frac{2}{\sqrt{1-x^2}}$. Differentiate ${y_2}$ with respect to $x$: $\frac{{d{y_2}}}{{dx}} = \frac{1}{2\sqrt{1-x^2}} 	imes (-2x) = -\frac{x}{\sqrt{1-x^2}}$. Therefore,the differential coefficient is $\frac{{d{y_1}}}{{d{y_2}}} = \frac{d{y_1}/dx}{d{y_2}/dx} = \frac{-2/\sqrt{1-x^2}}{-x/\sqrt{1-x^2}} = \frac{2}{x}$. At $x = \frac{1}{2}$,$\frac{{d{y_1}}}{{d{y_2}}} = \frac{2}{1/2} = 4$.

The differential coefficient of ${\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} \right)$ with respect to $\sqrt {1 - {x^2}} $ at $x = \frac{1}{2}$ is:

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