The problem asks to find the derivative of the function $f(x) = \text{sech}^{-1}(2x)$ at $x = 0.1$.

1. Problem Description

The problem asks to find the derivative of the function $f(x) = \text{sech}^{-1}(2x)$ at $x = 0.1$.

2. Solution Steps

First, we need to find the derivative of $f(x) = \text{sech}^{-1}(2x)$. The derivative of $\text{sech}^{-1}(u)$ is given by:

$\frac{d}{dx} \text{sech}^{-1}(u) = \frac{-1}{|u|\sqrt{1-u^2}} \frac{du}{dx}$

In our case, $u = 2x$. Thus, $\frac{du}{dx} = 2$. Plugging these into the formula, we get:

$f'(x) = \frac{-1}{|2x|\sqrt{1-(2x)^2}} \cdot 2 = \frac{-2}{|2x|\sqrt{1-4x^2}}$

Since we want to find $f'(0.1)$, we substitute $x = 0.1$:

$f'(0.1) = \frac{-2}{|2(0.1)|\sqrt{1-4(0.1)^2}} = \frac{-2}{|0.2|\sqrt{1-4(0.01)}} = \frac{-2}{0.2\sqrt{1-0.04}} = \frac{-2}{0.2\sqrt{0.96}}$

$f'(0.1) = \frac{-2}{0.2\sqrt{0.96}} = \frac{-10}{\sqrt{0.96}} \approx \frac{-10}{0.9798} \approx -10.2062$

Since the absolute value is in the denominator, it will always give us a positive number, but since the numerator is negative we keep that sign in front. The answer is around -10.

2

0

6

2. The question asks for the value, so we choose the option closest to this value.

3. Final Answer

B. 10.2062