Derivative of $\frac{d}{dx}\ln (x+\sqrt{x^2+1})$ Show that $\frac{d}{dx}\ln (x+\sqrt{x^2+1}) = \frac{1}{\sqrt{x^2+1}}$
So I've done this so far:
$$\frac{d}{dx}\ln (x+\sqrt{x^2+1}) = \frac{1+x(x^2+1)^{-0.5}}{x+(x^2+1)^{0.5}}$$
I have tried a few combinations but cannot attain the desired result. Could someone offer just a hint to get me going?
Thanks

Your differentiation is correct, and you applied the chain rule correctly. But as you note, your result can be simplified. 
Let's rewrite the fraction, expressing the term $x(x^2 + 1)^{-0.5}$ in the numerator as $\frac{x}{\sqrt {x^2 + 1}}$, then simply multiply the numerator and denominator by $\sqrt{x^2 + 1}$:
$$\begin{align}\frac{1+x(x^2+1)^{-0.5}}{x+(x^2+1)^{0.5}} & = \frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\\ \\
& = \frac{\sqrt{x^2 +1} + x}{{(x+\sqrt{x^2+1})\sqrt{x^2+1}}}\\ \\
& =\frac{\color{blue}{\bf x+\sqrt{x^2+1}}}{(\color{blue}{\bf x+\sqrt{x^2+1}})\sqrt{x^2+1}} \\ \\
& =\frac1{\sqrt{x^2+1}}\end{align}$$
