Chapter 3.7, Problem 45E

To determine

To calculate: The $n\text{th}$ derivative of the function $\ln \left(x-1\right)$ .

Answer to Problem 45E

The $n\text{th}$ derivative of the function $\ln \left(x-1\right)$ is $f^{\left(n\right)}\left(x\right)={\left(-1\right)}^{n+1}\frac{\left(n-1\right)!}{{\left(x-1\right)}^n}$ .

Explanation of Solution

Given information:

The function $\ln \left(x-1\right)$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Derivative of $\ln x$ is $\frac{1}{x}$ .

Quotient rule for differentiation is $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ where f and g are functions of x .

Calculation:

Consider the function $f\left(x\right)=\ln \left(x-1\right)$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\ln \left(x-1\right)\right)$

Recall that chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Derivative of $\ln x$ is $\frac{1}{x}$ .

Apply it.

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\ln \left(x-1\right)\right)\\ f\text{'}\left(x\right)=\frac{1}{\left(1-x\right)}\frac{d}{dx}\left(x-1\right)\\ f\text{'}\left(x\right)=\frac{1}{\left(x-1\right)}\left(1\right)\\ f\text{'}\left(x\right)=\frac{1}{\left(x-1\right)}\end{array}$

Again differentiate both sides with respect to x , apply the quotient rule for differentiation is $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ where f and g are functions of x .

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

Again differentiate both sides with respect to x , apply the quotient rule for differentiation is $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ where f and g are functions of x .

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

Now, observe that first derivative can be written as, $f\text{'}\left(x\right)={\left(-1\right)}^{1+1}\frac{\left(1-1\right)!}{{\left(x-1\right)}^1}$ .

Second derivative can be written as, $f\text{'}\text{'}\left(x\right)={\left(-1\right)}^{2+1}\frac{\left(2-1\right)!}{{\left(x-1\right)}^2}$ .

Third derivative can be written as, $f\text{'}\text{'}\text{'}\left(x\right)={\left(-1\right)}^{3+1}\frac{\left(3-1\right)!}{{\left(x-1\right)}^3}$ .

Similarly, $n\text{th}$ derivative is generalized as, $f^{\left(n\right)}\left(x\right)={\left(-1\right)}^{n+1}\frac{\left(n-1\right)!}{{\left(x-1\right)}^n}$

Thus, the $n\text{th}$ derivative of the function $\ln \left(x-1\right)$ is $f^{\left(n\right)}\left(x\right)={\left(-1\right)}^{n+1}\frac{\left(n-1\right)!}{{\left(x-1\right)}^n}$ .