Chapter 3.6, Problem 22E

To determine

To calculate: The derivative of the function $f\left(x\right)=x\ln \left(\arctan x\right)$ .

Answer to Problem 22E

The derivative of the function is $\frac{x}{\left(1+x^2\right){\tan }^{-1}x}+\ln \left({\tan }^{-1}x\right)$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=x\ln \left(\arctan x\right)$ .

Formula used:

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

Differentiation rule for inverse trigonometric functions is $\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$ .

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 .

Logarithmic rule for differentiation is $\frac{d}{dx}\ln x=\frac{1}{x}$ .

Calculation:

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

Differentiate both sides with respect to $x$ ,

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x\ln \left(\arctan x\right)\right)\\ \frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(x\ln \left(ta{n}^{-1}x\right)\right)\end{array}$

Recall that 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 and differentiation rule for inverse trigonometric functions is $\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$ .

Also apply logarithmic rule for differentiation and chain rule in the terms of the function.

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

Thus, the derivative of the function is $\frac{x}{\left(1+x^2\right){\tan }^{-1}x}+\ln \left({\tan }^{-1}x\right)$ .