Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do this?

My method:
$$
{f'(x)}=\left [\ln \left (\frac{x}{(x^2+1)^\frac{1}{2}} \right) \right ]'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right)\left (\frac{x}{(x^2+1)^\frac{1}{2}} \right)'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[(x^2+1)^\frac{1}{2}]'}{[(x^2+1)^\frac{1}{2}]^2} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(x^2+1)']}{\left | x^2+1 \right |} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)]}{x^2+1} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x^2(x^+1)^{-\frac{1}{2}}}{x^2+1} \right)=\frac{(x^2+1)^{(\frac{1}{2}+\frac{1}{2})}-x^2(x^2+1)^{\frac{1}{2}+-\frac{1}{2}{}}}{x(x^2+1)}=-\frac{x^2}{x}=-x
$$
and
$$
f''(x)=(-x)'=-1\
$$
This took me much more than 1.5 hours just to type into LaTex :'(

By implicit differentiation:
Let
$$
y(x) = \log\left[\frac{x}{\sqrt{x^2 + 1}}\right].
$$
Then
$$
(x^2 + 1)e^{2 y(x)} = x^2.
$$
Differentiating both sides,
$$
(x^2 + 1)e^{2y(x)}y'(x) + x e^{2y(x)} = x.
$$
Solving for $y'(x)$,
$$
y'(x) = \frac{x(e^{-2y(x)}-1)}{(x^2 + 1)} = \frac{1}{x(x^2+1)}.
$$
