CONTEXT: Question made up by uni lecturer
I need to show that $f(x)=\ln(x+\sqrt{x^2+1})$ is one-to-one using derivatives.
I know that the derivative of $f(x)$ can be found using $\frac{d}{dx}\ln g(x)=\frac{g'(x)}{g(x)}$.
For this question I found using the chain rule that $g'(x)=1+\frac{x}{\sqrt{x^2+1}}$.
Thus, $f'(x)=\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}$ which I simplified down to $f'(x)=\frac{x+\sqrt{x^2+1}}{x\sqrt{(x^2+1)^3}}$.
However, I've found using a derivative calculator that my answer for $f'(x)$ is incorrect and that $f'(x)=\frac{1}{\sqrt{x^2+1}}$.
Is my rule for deriving a natural log wrong, or did I derive $g(x)$ incorrectly, or did I do a simplification error? I know this is a pretty simple question but I've been stuck on it for ages and I don't know why.
How do I then show that $f'(x)>0$ algebraically?
Any help would be greatly appreciated.
The final (unexplained) simplification is incorrect, everything before it is fine.
From $\frac{1 + \frac{x}{\sqrt{x^2+1}}}{x + \sqrt {x^2+1}}$ you first take the common denominator on top to get $\frac{\left(\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}\right)}{x + \sqrt{x^2+1}}$, which now you write as $$\frac{(x+ \sqrt{x^2+1}) \times \left(\frac 1{\sqrt{x^2+1}}\right)}{x+\sqrt{x^2+1}} = \frac 1 {\sqrt{x^2+1}}$$
after cancellation. So you were fine up till then.
Indeed, I think you took the $\sqrt{x^2+1}$ to the bottom and got $\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}(x + \sqrt{x^2+1})}$, which is correct. From here cancellation works. But then you did something to the denominator to end up with $x\sqrt{(x^2+1)^3}$, which is not correct. So whatever has been done incorrectly has been at the denominator level.
Note that $f'(x) = \frac{1}{\sqrt{x^2+1}} > 0$ because both numerator and denominator are positive everywhere, and the ratio of positive quantities is positive.