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.