Chapter 3.7, Problem 23E

To determine

To calculate: The derivative and the domain of the function $f\left(x\right)=\frac{x}{1-\ln \left(x-1\right)}$ .

Answer to Problem 23E

The derivative of the function is $f\text{'}\left(x\right)=\frac{2x-1-\left(x-1\right)\ln \left(x-1\right)}{\left(x-1\right){\left(1-\ln \left(x-1\right)\right)}^2}$ and the domain of the function is $\left(1,1+e\right)\cup \left(1+e,\infty \right)$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{x}{1-\ln \left(x-1\right)}$ .

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 $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of 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 .

Domain of the function is the set of values of independent variable for which function is defined.

The function of the form $\frac{p}{q}$ fraction are defined when $q\ne 0$ that is denominator should not be equal to 0.

Calculation:

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

Differentiate both sides with respect to x ,

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

Recall that 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 .

Apply it.

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

Simplify it further as,

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

Thus, the derivative of the function is $f\text{'}\left(x\right)=\frac{2x-1-\left(x-1\right)\ln \left(x-1\right)}{\left(x-1\right){\left(1-\ln \left(x-1\right)\right)}^2}$ .

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

Recall that domain of the function is the set of values of independent variable for which function is defined.

The function of the form $\frac{p}{q}$ fraction are defined when $q\ne 0$ that is denominator should not be equal to 0.

The provided expression $f\left(x\right)=\frac{x}{1-\ln \left(x-1\right)}$ is defined for real values that is,

  $\begin{array}{c}1-\ln \left(x-1\right)\ne 0\\ \ln \left(x-1\right)\ne 1\\ x-1\ne e\\ x\ne 1+e\end{array}$

Also, logarithmic function is defined for positive values, that is,

  $\begin{array}{c}x-1>0\\ x>1\end{array}$

So, for the function to be defined $x\ne 1+e$ and $x>1$ .

Combine the above two inequalities and write in interval notation.

Therefore, domain is $\left(1,1+e\right)\cup \left(1+e,\infty \right)$ for function to be defined.

Hence, the domain of the function $f\left(x\right)=\frac{x}{1-\ln \left(x-1\right)}$ is $\left(1,1+e\right)\cup \left(1+e,\infty \right)$ .