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 :'(
 A: Hint
$$\ln(\frac{x}{\sqrt{x^2+1}})=\ln x-\frac{1}{2}\ln(x^2+1)$$
A: Take advantage of logarithm properties.
$$\ln\left(\frac{x}{\sqrt{x^2 + 1}}\right) = \ln(x) - \frac12\ln(x^2 + 1)$$
Then the derivative is easy:
$$f'(x) = \frac1x - \frac{x}{x^2 + 1}$$
A: The expression $$\frac{x}{\sqrt{x^2 + 1}}$$
is the sine of the angle adjacent to $1$, in a right triangle with legs of $1$ and $x$ and hypotenuse $\sqrt{x^2 + 1}$.
Therefore, it is equal to
$$
\sin \arctan x
$$
and we have
$$
\bbox[5px,border:2px solid blue]{f(x) =  \ln (\sin (\arctan x)).}
$$
Computing $f'(x)$ for this is just chain rule twice:
$$
f'(x) = \frac{1}{\sin \arctan x} \cdot \cos \arctan x \cdot \frac{1}{1 + x^2}.
$$
Now referring to the original triangle, $\sin \arctan x = \frac{x}{\sqrt{x^2 + 1}}$ and $\cos \arctan x = \frac{1}{\sqrt{x^2 + 1}}$, so
$$
f'(x) = \frac{\sqrt{x^2 + 1}}{x}
\cdot \frac{1}{\sqrt{x^2 + 1}}
\cdot \frac{1}{x^2 + 1}
= \frac{1}{x(x^2 + 1)}.
$$
Note your answer was incorrect.
A: Hint:
\begin{align}
f(x) = \log|x| - \frac{1}{2}\log|x^2+1|.
\end{align}
A: Let $f(x)=\log\left(\frac{x}{\sqrt{x^2+1}}\right)$ and 
$$ g(t)=f(\sinh t) = \log\tanh t = \log\sinh t-\log\cosh t. $$
We have
$$ \frac{d}{dt}g(t) = \frac{\cosh t}{\sinh t}-\frac{\sinh t}{\cosh t} = \frac{1}{\sinh(t)\cosh(t)}$$
but the LHS also equals $\cosh(t)\,f'(\sinh t)$, hence
$$ f'(\sinh t) = \frac{1}{\sinh(t)\cosh^2(t)} $$
and
$$ f'(x) = \color{red}{\frac{1}{x(1+x^2)}}.$$
We may freely assume $x>0$ since otherwise $f(x)$ makes no sense (and $f'(x)$ as well).
A: 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 $y'(x)$,
$$
y'(x) = \frac{x(e^{-2y(x)}-1)}{(x^2 + 1)} = \frac{1}{x(x^2+1)}.
$$
