Q. 39 - Calculus

Question

For each function f and value $x=c$ in Exercises 35–44, use a sequence of approximations to estimate $f\text{'}\left(c\right)$. Illustrate your work with an appropriate sequence of graphs of secant lines.

$f\left(x\right)=\ln \left(x^2+1\right), c=0$

Step-by-Step Solution

Answer

The value is $f\text{'}\left(0\right)=0$.  

Step 1. Given Information

We are given a function $f\left(x\right)=\ln \left(x^2+1\right), c=0$ and using a sequence of approximations we have to estimate $f\text{'}\left(c\right)$.  

Step 2. Using a sequence of approximations

$$\begin{gathered} h=1, \\ \frac{f\left(1\right)-f\left(0\right)}{1-0} \\ =\frac{\ln \left({\left(1\right)}^2+1\right)-\ln \left({\left(0\right)}^2+1\right)}{1} \\ =\frac{\ln \left(2\right)-0}{1} \\ =\ln \left(2\right) \\ =0.6931 \\ h=0.5, \\ \frac{f(0.5)-f\left(0\right)}{0.5-0} \\ =\frac{\ln \left({\left(0.5\right)}^2+1\right)-\ln \left({\left(0\right)}^2+1\right)}{0.5} \\ =\frac{\ln \left(1.25\right)-0}{0.5} \\ =0.4462 \\ h=0.1, \\ \frac{f(0.1)-f\left(0\right)}{0.1-0} \\ =\frac{\ln \left({\left(0.1\right)}^2+1\right)-\ln \left({\left(0\right)}^2+1\right)}{0.1} \\ =\frac{\ln \left(1.01\right)-0}{0.1} \\ =0.0995 \\ h=0.01, \\ \frac{f(0.01)-f\left(0\right)}{0.01-0} \\ =\frac{\ln \left({\left(0.01\right)}^2+1\right)-\ln \left({\left(0\right)}^2+1\right)}{0.01} \\ =\frac{\ln \left(1.0001\right)-0}{0.01} \\ =0.009 \end{gathered}$$

We might guess that the slope of the tangent line is $f\text{'}\left(0\right)=0$.

The graph of the function is 

Step 3. Graphs of secant lines

When $c=0, c+h=1$ with $h=1$ we have $f\left(0\right)=, f\left(1\right)=\ln \left(2\right)$ the secant line can be drawn as follows:

When $c=0, c+h=0.5$ with $h=0.5$ we have $f\left(0\right)=0, f(0.5)=0.2231$ the secant line can be drawn as follows:

When $c=0, c+h=0.1$ with $h=0.1$ we have $f\left(0\right)=0, f(0.1)=0.0099$ the secant line can be drawn as follows: