Simplifying Derivatives with the Chain Rule

[1MileCrash]

Homework Statement

$f(x) = ln(x+\sqrt{x^2+1})$

Homework Equations

The Attempt at a Solution

First, I applied the chain rule.

$[\frac{1}{x+\sqrt{x^2+1}}]Dx[x+\sqrt{x^2+1}]$

Second, to find $Dx[x+\sqrt{x^2+1}]$, I broke it into two derivatives. Derivative of x is 1, so

$1 + Dx[\sqrt{x^2+1}]$

To find $Dx[\sqrt{x^2+1}]$, I applied the chain rule once more.

$[\frac{1}{2}][2x]\frac{1}{\sqrt{x^2+1}}$

I simplified this result to:

$\frac{x}{\sqrt{x^2+1}}$


Leading to and end-derivative of:

$[\frac{1}{x+\sqrt{x^2+1}}][1+\frac{x}{\sqrt{x^2+1}}]$

The book gives a much cleaner answer of $\frac{1}{\sqrt{x^2+1}}$

Is my answer equivalent? If yes, how would I get to that? If no, what part of the calculus did I screw up?


WOW, Nevermind!

 

[SammyS]

Do some algebra.

Find a common denominator for $\displaystyle 1+\frac{x}{\sqrt{x^2+1}}\,.$ & write as one fraction.