Find the derivative: frac{d}{dx} cosh^{-1}(se | vedclass

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(A) We are given the function $y = \cosh^{-1}(\sec x)$.Using the chain rule,the derivative is $\frac{dy}{dx} = \frac{d}{dx} \cosh^{-1}(\sec x)$.The de

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$\sec x$

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(A) We are given the function $y = \cosh^{-1}(\sec x)$.Using the chain rule,the derivative is $\frac{dy}{dx} = \frac{d}{dx} \cosh^{-1}(\sec x)$.The derivative of $\cosh^{-1}(u)$ is $\frac{1}{\sqrt{u^2 - 1}} \cdot \frac{du}{dx}$.Here,$u = \sec x$,so $\frac{du}{dx} = \sec x 	an x$.Substituting these into the formula:$\frac{dy}{dx} = \frac{1}{\sqrt{\sec^2 x - 1}} \cdot (\sec x 	an x)$.Since $\sec^2 x - 1 = 	an^2 x$,we have $\sqrt{\sec^2 x - 1} = 	an x$ (assuming $	an x > 0$ for the domain).Therefore,$\frac{dy}{dx} = \frac{1}{	an x} \cdot \sec x 	an x = \sec x$.

Find the derivative: $\frac{d}{dx} \cosh^{-1}(\sec x) = $

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