Question

Use logarithmic differentiation to find derivative

$f\left(x\right)=(\sec x{)}^x$

Short answer

The answer of this question is

$f^{\mathrm{\prime }}\left(x\right)=(\ln (\sec x)+x\tan x)(\sec x{)}^x$

Step-by-step solution

Step 1.Given information

We have been given to find the derivative of function

$f\left(x\right)=(\sec x{)}^x$

Step 2.Derivative of given function 

Product rule

$\left(fg{)}^{\mathrm{\prime }}\right(x)=f^{\mathrm{\prime }}(x\left)g\right(x)+f(x\left)g^{\mathrm{\prime }}\right(x)$

Quotient rule

${\left(\frac{f}{g}\right)}^{\mathrm{\prime }}\left(x\right)=\frac{f^{\mathrm{\prime }}\left(x\right)g\left(x\right)-f\left(x\right)g^{\mathrm{\prime }}\left(x\right)}{\left(g\right(x){)}^2}$

Step 3.Using logarithmic approach 

Taking log both side and then approaching

$\left.\begin{array}{r}f\left(x\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=(\sec x{)}^x\\ \ln f\left(x\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\ln (\sec x{)}^x\text{[Taking log in both sides]}\\ \ln f\left(x\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=x\ln (\sec x)\\ \frac{f^{\mathrm{\prime }}\left(x\right)}{f\left(x\right)}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{d}{dx}(x\ln (\sec x\left)\right)\left[\begin{array}{c}\text{Differentiate both sides}\\ \text{with respect to}x\end{array}\right]\\ \frac{f^{\mathrm{\prime }}\left(x\right)}{f\left(x\right)}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\left(\frac{d}{dx}\left(x\right)\right)\ln (\sec x)+x\left(\frac{d}{dx}\ln (\sec x)\right)\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\ln (\sec x)+\frac{x}{\sec x}\left(\frac{d}{dx}\sec x\right)\left[\begin{array}{c}\text{Using product rule}\\ \frac{d}{dx}\ln x=\frac{1}{x}\end{array}\right]\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\ln (\sec x)+\frac{x}{\sec x}(\sec x\tan x)\\ f^{\mathrm{\prime }}\left(x\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\left(\frac{d}{dx}\ln (\sec x=\sec x\tan x]\right.\end{array}\right]$