[SOLVED] Deriving a square root under a square

**pseizure2000** · September 28th 2007, 05:54 PM

Suppose that

. Find

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

**janvdl** · September 29th 2007, 12:27 AM

Originally Posted by pseizure2000

Suppose that

. Find

(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.

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

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

**ticbol** · September 29th 2007, 02:49 AM

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.

**topsquark** · September 29th 2007, 03:00 AM

Originally Posted by pseizure2000

Suppose that

. Find

(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

**Krizalid** · September 29th 2007, 06:21 AM

Mmm... see here.

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