Question

Find the derivative. Simplify where possible.

49. \(y = {\cosh ^{ - 1}}\left( {\sec \theta } \right)\)

Short answer

The derivative is \(y' = \sec \theta \).

Step-by-step solution

Apply the derivative of composite function

On applying the derivative of the given function\(y = {\cosh ^{ - 1}}\left( {\sec \theta } \right)\), we get, as follows:

^.\(y' = \frac{d}{{d\theta }}\left( {{{\cosh }^{ - 1}}\left( {\sec \theta } \right)} \right)\frac{d}{{d\theta }}\left( {\sec \theta } \right)\)

Plug in the derivatives and simplify

The derivative of\({\cosh ^{ - 1}}\theta \)is \(\frac{1}{{\sqrt {{\theta ^2} - 1} }}\) and that of \(\sec \theta \) is \(\sec \theta \tan \theta \).On plugging this derivative and simplifying we get:

\(\begin{aligned}y' & = \frac{d}{{d\theta }}\left( {{{\cosh }^{ - 1}}\left( {\sec \theta } \right)} \right)\frac{d}{{d\theta }}\left( {\sec \theta } \right)\\ & = \frac{1}{{\sqrt {{{\sec }^2}\theta - 1} }} \cdot \sec \theta \tan \theta \\ & = \frac{1}{{\sqrt {{{\tan }^2}\theta } }} \cdot \sec \theta \tan \theta \\ & = \frac{1}{{\tan \theta }} \cdot \sec \theta \tan \theta \\ & = \sec \theta \end{aligned}\)

Thus, \(y' = \sec \theta \).