Chapter 7.2, Problem 9QR

To determine

  $\text{ To find }\frac{dy}{dx}\text{.}$

Answer to Problem 9QR

  $\text{The required solution is, }\frac{dy}{dx}=\sec x.$

Explanation of Solution

Given information: $\text{The equation is},y=\ln \left(\sec x+\tan x\right)$

Calculation : $\frac{dy}{dx}\text{ is obtained as,}$

  $\begin{array}{l}\text{Find the derivative using chain rule,}\\ y=\ln \left(\sec x+\tan x\right)\\ \frac{dy}{dx}=\frac{1}{\sec x+\tan x}\text{.}\frac{d}{dx}\left(\sec x+\tan x\right)\\ =\frac{1}{\sec x+\tan x}\text{.}\left(\sec x\tan x+{\sec }^{2}x\right)\\ =\frac{1}{\sec x+\tan x}\text{.}\sec x\left(\tan x+\sec x\right)\\ =\sec x\\ \text{Hence, the required solution is, }\frac{dy}{dx}=\sec x.\end{array}$