# [SOLVED] Deriving a square root under a square

• Sep 28th 2007, 06:54 PM
pseizure2000
[SOLVED] Deriving a square root under a square
Suppose that http://www.mathhelpforum.com/math-he...6bb0913c-1.gif. Find http://www.mathhelpforum.com/math-he...10b0a6cb-1.gif

(moved from urgent and fixed) oh man I really need help on this one.
• Sep 29th 2007, 01:27 AM
janvdl
Quote:

Originally Posted by pseizure2000
Suppose that http://www.mathhelpforum.com/math-he...6bb0913c-1.gif. Find http://www.mathhelpforum.com/math-he...10b0a6cb-1.gif

(moved from urgent and fixed) oh man I really need help on this one.

I know the thread says solved, but i'll do it anyway so you can check.

$f(x) = (6 + (8x)^{ \frac{1}{2} } )^{ \frac{1}{2} }$
$f(x) = 6^{ \frac{1}{2} } + (8x)^{ \frac{1}{4} }$
$f(x) = 6^{ \frac{1}{2} } + (8)^{ \frac{1}{4} } \times (x)^{ \frac{1}{4} }$

$f '(x) = \frac{ \frac{1}{4} \sqrt[4]{8} }{ \sqrt[4]{x^3} }$
• Sep 29th 2007, 03:49 AM
ticbol
Quote:

Originally Posted by janvdl
I know the thread says solved, but i'll do it anyway so you can check.

$f(x) = (6 + (8x)^{ \frac{1}{2} } )^{ \frac{1}{2} }$
$f(x) = 6^{ \frac{1}{2} } + (8x)^{ \frac{1}{4} }$
$f(x) = 6^{ \frac{1}{2} } + (8)^{ \frac{1}{4} } \times (x)^{ \frac{1}{4} }$

$f '(x) = \frac{ \frac{1}{4} \sqrt[4]{8} }{ \sqrt[4]{x^3} }$

But (a +b)^(1/2) is not equal to a^(1/2) +b^(1/2)

Or more specifically,
(6 +(8x)^(1/2))^(1/2)
is not
6^(1/2) +(8x)^(1/4)

Example,
(4 +sqrt(81))^(1/2)
is not
4^(1/2) +(81)^(1/4) = 2 +3 = 5

(4 +sqrt(81))^(1/2)
= (4 +9)^(1/2)
= sqrt(13) ----------------------not 5.
• Sep 29th 2007, 04:00 AM
topsquark
Quote:

Originally Posted by pseizure2000
Suppose that http://www.mathhelpforum.com/math-he...6bb0913c-1.gif. Find http://www.mathhelpforum.com/math-he...10b0a6cb-1.gif

(moved from urgent and fixed) oh man I really need help on this one.

$f(x) = \sqrt{6 + \sqrt{8x}}$

Write this as:
$f(x) = (6 + (8x)^{1/2} )^{1/2}$

Now use the chain rule:
$f^{\prime}(x) = \frac{1}{2} \cdot (6 + (8x)^{1/2})^{-1/2} \cdot \left ( \frac{1}{2} \cdot (8x)^{-1/2} \cdot 8 \right )$

$f^{\prime}(x) = \frac{2}{\sqrt{6 + \sqrt{8x}}\sqrt{8x}}$

-Dan
• Sep 29th 2007, 07:21 AM
Krizalid
Mmm... see here.