Answer::
[tex]f^{\prime}(x)=7x\sec x\tan x+7\sec x+\frac{1}{x}[/tex]Explanation:
Given f(x) defined below:
[tex]f(x)=\ln x+7x\sec x[/tex]The derivative is calculated below.
[tex]\begin{gathered} \frac{d}{dx}\lbrack f(x)\rbrack=\frac{d}{dx}\lbrack\ln x+7x\sec x\rbrack \\ =\frac{d}{dx}\lbrack\ln x\rbrack+\frac{d}{dx}\lbrack7x\sec x\rbrack \\ Take\text{ the constant 7 outside the derivative sign.} \\ =$\textcolor{red}{\frac{d}{dx}\lbrack\ln x\rbrack}$+7\frac{d}{dx}\lbrack x\sec x\rbrack \\ \text{The derivative of }\ln (x)=\frac{1}{x},\text{ therefore:} \\ $\textcolor{red}{\frac{d}{dx}\lbrack\ln x\rbrack}$+7\frac{d}{dx}\lbrack x\sec x\rbrack=$\textcolor{red}{\frac{1}{x}}$+7\frac{d}{dx}\lbrack x\sec x\rbrack\cdots(1) \end{gathered}[/tex]Next, we find the derivative of x sec x using the product rule.
[tex]\begin{gathered} \frac{d}{dx}\lbrack x\sec x\rbrack=x$\textcolor{blue}{\frac{d}{dx}\lbrack\sec x\rbrack}$+\sec x\frac{d}{dx}\lbrack x\rbrack\text{ } \\ The\text{ derivative of sec(x), }\text{\textcolor{red}{ }}\textcolor{red}{\frac{d}{dx}\lbrack\sec x\rbrack=\sec x\tan x} \\ =x$\textcolor{blue}{\lbrack\sec x\tan x\rbrack}$+\sec x \end{gathered}[/tex]Substitute the result into equation (1) above.
[tex]\begin{gathered} \frac{1}{x}+7\frac{d}{dx}\lbrack x\sec x\rbrack=\frac{1}{x}+7(x\sec x\tan x+\sec x) \\ =7x\sec x\tan x+7\sec x+\frac{1}{x} \end{gathered}[/tex]Therefore:
[tex]f^{\prime}(x)=7x\sec x\tan x+7\sec x+\frac{1}{x}[/tex]