Chapter 10, Problem 1RE

To determine

To find: The fourth derivative of the function $f\left(x\right)=\ln \left(x+5\right)$.

Answer to Problem 1RE

The fourth derivative of the function $f\left(x\right)=\ln \left(x+5\right)$ is $f^4\left(x\right)=-\frac{6}{{\left(x+5\right)}^4}$.

Explanation of Solution

Obtain the first derivatives of $f\left(x\right)=\ln \left(x+5\right)$ is,

$\begin{array}{c}{f}^1\left(x\right)=\frac{d}{dx}\ln \left(x+5\right)\\ =\frac{1}{x+5}\frac{d}{dx}\left(x+5\right)\\ =\frac{1}{x+5}\left(1\right)\\ =\frac{1}{x+5}\end{array}$

Obtain the second derivatives of $f\left(x\right)=\ln \left(x+5\right)$ is,

$\begin{array}{c}{f}^2\left(x\right)=\frac{d}{dx}\left(\frac{1}{x+5}\right)\\ =\frac{\left(x+5\right)\frac{d}{dx}\left(1\right)-\left(1\right)\frac{d}{dx}\left(x+5\right)}{{\left(x+5\right)}^2}\\ =\frac{0-1}{{\left(x+5\right)}^2}\\ =\frac{-1}{{\left(x+5\right)}^2}\end{array}$

Obtain the third derivatives of $f\left(x\right)=\ln \left(x+5\right)$ is,

$\begin{array}{c}{f}^3\left(x\right)=\frac{d}{dx}\left(\frac{-1}{{\left(x+5\right)}^2}\right)\\ =\frac{{\left(x+5\right)}^2\frac{d}{dx}\left(-1\right)-\left(-1\right)\frac{d}{dx}\left({\left(x+5\right)}^2\right)}{{\left({\left(x+5\right)}^2\right)}^2}\\ =\frac{0+2\left(x+5\right)}{{\left(x+5\right)}^4}\\ =\frac{2}{{\left(x+5\right)}^4}\end{array}$

Obtain the fourth derivatives of $f\left(x\right)=\ln \left(x+5\right)$ is,

$\begin{array}{c}{f}^4\left(x\right)=\frac{d}{dx}\left(\frac{2}{{\left(x+5\right)}^3}\right)\\ =\frac{{\left(x+5\right)}^3\frac{d}{dx}\left(2\right)-\left(2\right)\frac{d}{dx}{\left(x+5\right)}^3}{{\left({\left(x+5\right)}^3\right)}^2}\\ =\frac{0-6{\left(x+5\right)}^2}{{\left(x+5\right)}^6}\\ =\frac{-6}{{\left(x+5\right)}^4}\end{array}$

Therefore, the fourth derivative of the function $f\left(x\right)=\ln \left(x+5\right)$. is $f^4\left(x\right)=-\frac{6}{{\left(x+5\right)}^4}$.