Question

Consider the function

$F\left(x\right)=\left\{\begin{array}{l}se{c}^{-1}x,-\mathrm{\pi }\mathrm{if}\mathrm{x}<-1\\ se{c}^{-1}x+\mathrm{\pi }\mathrm{if}\mathrm{x}>1\end{array}\right.$

Show that the derivative of this function is the function $f\left(x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}}$. Compare the graphs of $F\left(x\right)$and$se{c}^{-1}x$, and discuss how this exercise relates to the second part of Theorem 4.19

Short answer

Proved that $F\text{'}\left(x\right)=\frac{x}{\left|x\right|\left|x\right|\sqrt{{\left|x\right|}^2-1}},x>1andx<-1$and graph as show below

Step-by-step solution

Step 1. Given information

The given function$F\left(x\right)=\left\{\begin{array}{l}se{c}^{-1}x,-\mathrm{\pi }\mathrm{if}\mathrm{x}<-1\\ se{c}^{-1}x+\mathrm{\pi }\mathrm{if}\mathrm{x}>1\end{array}\right.$

Step 2. Redefine given function using graph of the function sec-1(x)

The graph of the functions $se{c}^{-1}\left(x\right),se{c}^{-1}\left(x\right)-\mathrm{\pi }$and $se{c}^{-1}\left(x\right)+\mathrm{\pi }$are shown in the following figure.

From the graph,it is clear that the function $F\left(x\right)$may be written as $F\left(x\right)=se{c}^{-1}\left|x\right|,x>1$and $x<-1$or $\left|x\right|>1$

Further,let

$$\begin{gathered} \left|x\right|=y \\ \Rightarrow F\left(x\right)=se{c}^{-1}y \end{gathered}$$.

Therefore,$y=secF$

$\Rightarrow \frac{dy}{dx}=\left(secF\tan F\right)F\text{'}\left(x\right)...............\left(1\right)$

We have $$\begin{gathered} y=\left|x\right| \\ \Rightarrow \frac{dy}{dx}=\frac{x}{\left|x\right|}.....................\left(2\right) \end{gathered}$$

From results (1) and (2) we got $$\begin{gathered} \left(secF\tan F\right)F\text{'}\left(x\right)=\frac{x}{\left|x\right|} \\ F\text{'}\left(x\right)=\frac{x}{\left|x\right|}\frac{1}{\left(secF\tan F\right)}...............\left(3\right) \end{gathered}$$

$secF=\frac{y}{1}$,Therefore,from reference triangle

$\tan F=\sqrt{y^2-1}$

Substitute the values in equation (3)

$$\begin{gathered} F\text{'}\left(x\right)=\frac{x}{\left|x\right|}\frac{1}{y\sqrt{y^2-1}} \\ =\frac{x}{\left|x\right|\left|x\right|\sqrt{{\left|x\right|}^2-1}},x>1andx<-1 \end{gathered}$$

Hence,proved .

Step 3.  Draw the graph for sec-1x

Now,$F\text{'}\left(x\right)=f\left(x\right)$that implies $F\left(x\right)$is antiderivative of function$f\left(x\right)$

Hence,

$$\begin{gathered} F\left(x\right)=\int f\left(x\right)dx+C \\ =\int \frac{1}{\left|x\right|\sqrt{x^2-1}}dx+C \\ =se{c}^{-1}x+C \end{gathered}$$