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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\sec x$

Vedclass · Answer

(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) = $

Similar Questions

If $y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right)$,then $y'(1)$ is

$\frac{d}{dx} \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} = $

If $y=\log \left[\tan \sqrt{\frac{2^x-1}{2^x+1}}\right], x>0$,then $\left(\frac{d y}{d x}\right)_{x=1}=$

The derivative of $\cosh^{-1} x$ with respect to $\log x$ at $x=5$ is

If $f(x) = e^x g(x)$,$g(0) = 2$,and $g'(0) = 1$,then $f'(0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module