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

In summary, the conversation was discussing a problem involving finding the derivative y' using a given equation. The main issue was understanding the simplification from equation (2) to (3), which involved multiplying and dividing by the square root of the denominator. The conversation also touched on using the fact that x = (\sqrt{x})^2 = \sqrt{x}\cdot\sqrt{x} to help understand the simplification.
  • #1
5hassay
82
0

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).

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

Homework Equations



If it helps, [itex]y = \sqrt{x + \sqrt{x^{2} + 1}}[/itex]

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!
 
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  • #2
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
Multiply top and bottom by [tex]\frac{\frac{1}{\sqrt{x+\sqrt{x^2+1}}}}

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

Pretty much what Watermelonpig said.
 
  • #4
Using the fact that [itex]x = (\sqrt{x})^2 = \sqrt{x}\cdot\sqrt{x}[/itex], try rewriting the top term in the brackets in (2) as
[itex]\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}}[/itex]
 
  • #5
Ah! Thank you very much WatermelonPig, gb7nash, and Bohrok -- I do understand now.
 

What is the formula for simplifying the derivative of the square root of a sum containing a square root?

The formula is: d/dx √(a + √b) = (√(4a + 2√b))/(2√(a + √b))

What are the steps for simplifying the derivative of the square root of a sum containing a square root?

The steps are: 1. Rewrite the expression as (√(a + √b))/(√(a + √b))2. Multiply the expression by (√(a + √b))/(√(a + √b))3. Simplify the numerator and denominator using the distributive property4. Simplify the expression by combining like terms and taking out common factors5. Simplify the final expression by dividing both the numerator and denominator by 2

How do you simplify the derivative of the square root of a sum containing a square root with variables?

To simplify the derivative, follow the same steps as above, but instead of using numeric values, leave the variables in the expression. For example, d/dx √(x + √y) = (√(4x + 2√y))/(2√(x + √y))

What is the purpose of simplifying the derivative of the square root of a sum containing a square root?

The purpose of simplifying the derivative is to make it easier to find the slope of a curve at a specific point. This can be useful in various applications, such as finding maximum and minimum values, determining rates of change, and solving optimization problems.

Are there any special cases to consider when simplifying the derivative of the square root of a sum containing a square root?

Yes, if the expression inside the square root is a perfect square, the simplification process becomes simpler. In this case, the final expression will have no square roots and can be simplified further. Also, if the expression inside the square root is negative, the derivative will not exist.

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