How the right term of the derivative is gained? This deduction is one of the typical ones I think.
What I want to deduce is the right term from the left term of the below equation.
$$\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)=\frac{-a}{x\sqrt{a^{2}+x^{2}}}$$
$\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\frac{d}{dx}\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\frac{d}{dx}\left(x^{-1}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((x^{-1})'\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((-1\cdot x^{-2})\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left(-x^{-2}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(\left(a^{2}+x^{2}\right)^{\frac{1}{2}}\right)\right)$
Anyone deduced it in someday?
ps. I have to go to work. Back after about 7hours.
 A: Small error: a missing + sign in the fourth line from the product rule:
$$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((x^{-1})'\left(a+\sqrt{a^{2}+x^{2}}\right) ~~\mathbf{+}~~ x^{-1}\frac{d}{dx}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$$
Continuing on with your idea, the next step would be to use the chain rule on that square root; with the + sign in the proper place we get
$$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right) \left(-\frac{1}{x^2}\left(a+\sqrt{a^{2}+x^{2}}\right) ~~\mathbf{+}~~ \frac{1}{x}\frac{x}{\sqrt{a^{2}+x^{2}}}\right)$$
Then the main trick is to get a common denominator:
$$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right) \frac{-a\sqrt{a^2+x^2}-a^2-x^2 + x^2}{x^2\sqrt{a^2+x^2}}$$
and I'm sure you can take it from here.
A: $$f(x)=\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)=\frac{d}{dx}\left(\log\left(a+\sqrt{a^{2}+x^{2}}\right)\right)-\frac 1x$$
$$\frac{d}{dx}\left(\log\left(a+\sqrt{a^{2}+x^{2}}\right)\right)=\frac{\Big[a+\sqrt{a^{2}+x^{2}}\Big]' }{ a+\sqrt{a^{2}+x^{2}}}$$
$$\Big[a+\sqrt{a^{2}+x^{2}}\Big]'=\frac{x}{\sqrt{a^2+x^2}}$$
$$f(x)=\frac{x}{\sqrt{a^2+x^2} \left(\sqrt{a^2+x^2}+a\right)}-\frac{1}{x}$$
$$f(x)=\frac{x\left(\sqrt{a^2+x^2}-a\right)}{\sqrt{a^2+x^2} \left(\sqrt{a^2+x^2}+a\right)\left(\sqrt{a^2+x^2}-a\right)}-\frac{1}{x}$$
Just finish.
