Question

Find the derivative. Simplify where possible.

50. \(y = {{\mathop{\rm sech}\nolimits} ^{ - 1}}\left( {\sin \theta } \right)\)

Short answer

The derivative is \(y' = - \csc \theta \).

Step-by-step solution

Apply the derivative of composite function

On applying the derivative of the given function\(y = {{\mathop{\rm sech}\nolimits} ^{ - 1}}\left( {\sin \theta } \right)\), we get, as follows:

^.\(y' = \frac{d}{{d\theta }}\left( {{{{\mathop{\rm sech}\nolimits} }^{ - 1}}\left( {\sin \theta } \right)} \right)\frac{d}{{d\theta }}\left( {\sin \theta } \right)\)

Plug in the derivatives and simplify

The derivative of\({{\mathop{\rm sech}\nolimits} ^{ - 1}}\theta \)is \( - \frac{1}{{\theta \sqrt {1 - {\theta ^2}} }}\) and that of \(\sin \theta \) is \(\cos \theta \).On plugging this derivative and simplifying we get:

\(\begin{aligned}y' & = \frac{d}{{d\theta }}\left( {{{{\mathop{\rm sech}\nolimits} }^{ - 1}}\left( {\sin \theta } \right)} \right)\frac{d}{{d\theta }}\left( {\sin \theta } \right)\\ & = - \frac{1}{{\sin \theta \sqrt {1 - {{\sin }^2}\theta } }}\left( {\cos \theta } \right)\\ & = - \frac{{\cos \theta }}{{\sin \theta \cos \theta }}\\ & = - \frac{1}{{\sin \theta }}\\ & = - \csc \theta \end{aligned}\)

So, \(y' = - \csc \theta \).