Simplifying the Derivative of the Square Root of a Sum Containing a Square Root

[5hassay]

Homework Statement

EDIT: Ahhh, my apologies. At first, I thought it appropriate for the non-calculus sub-forum, but by the title it really does not, XD. I also can't seem to find how to remove it. The question really does not require calculus, though!

Basically, the problem of finding the derivative y' is fine, but there is a point at which the text further simplifies the derivative in a method I do not understand, specifically from equation (2) to (3).

D_{x}y = \frac{1}{2\sqrt{x + \sqrt{x^{2}+1}}}\left[1 + \frac{x}{\sqrt{x^{2}+1}}\right] (1)
D_{x}y = \frac{1}{2\sqrt{x + \sqrt{x^{2}+1}}}\left[\frac{\sqrt{x^{2}+1} + x}{\sqrt{x^{2}+1}}\right] (2)
D_{x}y = \frac{\sqrt{x + \sqrt{x^{2}+1}}}{2 \sqrt{x^{2}+1}} (3)

Homework Equations

If it helps, y = \sqrt{x + \sqrt{x^{2} + 1}}

The Attempt at a Solution

I have tried a few things, such as multiplying and dividing (2) by the numerator of (3), adding both of the squares in the denominators of (2), and various other attempts, such as trying to go from (3) to (2) or (1). However, I don't seem to get anywhere. What is this silly small thing I am probably not seeing, XD?

Much appreciation for any help!

 

[WatermelonPig]

Something divided by the square root of itself is the square root of itself. Which in this case is x + sqrt(x^2+1).

 

[gb7nash]

Multiply top and bottom by \frac{\frac{1}{\sqrt{x+\sqrt{x^2+1}}}}<br /> <br /> {\frac{1}{\sqrt{x+\sqrt{x^2+1}}}}

Pretty much what Watermelonpig said.

 

[Bohrok]

Using the fact that x = (\sqrt{x})^2 = \sqrt{x}\cdot\sqrt{x}, try rewriting the top term in the brackets in (2) as
\sqrt{x^2 + 1} + x = (\sqrt{x + \sqrt{x^2 + 1}})^2 = \sqrt{x + \sqrt{x^2 + 1}} \cdot \sqrt{x + \sqrt{x^2 + 1}}

 

[5hassay]

Ah! Thank you very much WatermelonPig, gb7nash, and Bohrok -- I do understand now.