# Error in finding derivative of a natural log to show function is 1-1

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.

• Ahh yes I did take the $\sqrt{x^2+1}$ to the bottom. Your explanation makes so much sense, thanks so much! – Ruby Pa Mar 17 at 8:07
• You are welcome! – астон вілла олоф мэллбэрг Mar 17 at 8:08