Question

Find \(y'\) and \(y''\).

31. \(y = \ln \left| {\sec x} \right|\)

Short answer

The values of \(y'\) and \(y''\) is \(y' = \tan x\) and \(y'' = {\sec ^2}x\).

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunction is shown below:

1. \(\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}\)
2. \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
3. \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) or \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Evaluate the values of  \(y'\) and \(y''\)

Evaluate the first derivative of the function as shown below:

\(\begin{aligned}{c}y'&= \frac{d}{{dx}}\left( {\ln \left| {\sec x} \right|} \right)\\&= \frac{1}{{\sec x}} \cdot \frac{d}{{dx}}\left( {\sec x} \right)\\&= \frac{1}{{\sec x}} \cdot \left( {\sec x\tan x} \right)\\&= \tan x\end{aligned}\)

The value of \(y'\) is \(y' = \tan x\).

Evaluate the second derivative of the function as shown below:

\(\begin{aligned}{c}y''&= \frac{d}{{dx}}\left( {\tan x} \right)\\&= {\sec ^2}x\end{aligned}\)

The value of \(y''\) is \(y'' = {\sec ^2}x\).

Thus, the values of \(y'\) and \(y''\) is \(\tan x\) and \({\sec ^2}x\).