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

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Question

(C) The derivative of the inverse trigonometric function $\sec^{-1} x$ with respect to $x$ is a standard formula in calculus.By definition,if $y = \se

Vedclass · Accepted Answer

$\frac{1}{x\sqrt{x^2-1}}$

Vedclass · Answer

(C) The derivative of the inverse trigonometric function $\sec^{-1} x$ with respect to $x$ is a standard formula in calculus.By definition,if $y = \sec^{-1} x$,then $x = \sec y$.Differentiating both sides with respect to $y$,we get $\frac{dx}{dy} = \sec y 	an y$.Since $	an y = \sqrt{\sec^2 y - 1} = \sqrt{x^2 - 1}$,we have $\frac{dx}{dy} = x\sqrt{x^2 - 1}$.Therefore,$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{x\sqrt{x^2 - 1}}$.Thus,the correct option is $C$.

The differential coefficient of $\sec^{-1} x$ is:

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