Given the function : Find the derivative of the function with respect to . An analyst claims that…

We simplify $f(x)$ before differentiating: Rationalizing the argument of the logarithm: $\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}-x} = \frac{(\sqrt{1+x^2}+x)^2}{(1+x^2) - x^2} = (\sqrt{1+x^2}+x)^2$. Thus, $\log\left[\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}-x}\right] = \log(\sqrt{1+x^2}+x)^2 = 2\log(\sqrt{1+x^2}+x)$. Also, $\tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x$. So $f(x) = 2\log(\sqrt{1+x^2}+x) + 2\tan^{-1}x$. Differentiating with respect to $x$: $\frac{d}{dx}[2\log(\sqrt{1+x^2}+x)] = 2 \cdot \frac{1}{\sqrt{1+x^2}+x} \cdot \left(\frac{2x}{2\sqrt{1+x^2}} + 1\right) = 2 \cdot \frac{1}{\sqrt{1+x^2}+x} \cdot \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}} = \frac{2}{\sqrt{1+x^2}}$. $\frac{d}{dx}[2\tan^{-1}x] = \frac{2}{1+x^2}$. Therefore, $f'(x) = \frac{2}{\sqrt{1+x^2}} + \frac{2}{1+x^2}$.

The analyst's claim is incorrect. For all real $x$, $x^2 \ge 0$, so $\sqrt{1+x^2} \ge 1$ and $1+x^2 \ge 1$. Thus, $\frac{2}{\sqrt{1+x^2}} > 0$ and $\frac{2}{1+x^2} > 0$, which means $f'(x) > 0$ for all $x \in \mathbb{R}$. Since $f'(x)$ is strictly positive everywhere, $f(x)$ is strictly increasing on $\mathbb{R}$ and $f'(x)$ never equals zero. Therefore, $f(x)$ has no local extrema (neither maximum nor minimum).