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

1. Jul 22, 2011

### 5hassay

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

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

3. 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!

2. Jul 22, 2011

### 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).

3. Jul 22, 2011

### gb7nash

Multiply top and bottom by $$\frac{\frac{1}{\sqrt{x+\sqrt{x^2+1}}}} {\frac{1}{\sqrt{x+\sqrt{x^2+1}}}}$$

Pretty much what Watermelonpig said.

4. Jul 22, 2011

### 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
$\sqrt{x^2 + 1} + x = (\sqrt{x + \sqrt{x^2 + 1}})^2 = \sqrt{x + \sqrt{x^2 + 1}} \cdot \sqrt{x + \sqrt{x^2 + 1}}$

5. Jul 25, 2011

### 5hassay

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