# derivative question

• Mar 4th 2010, 07:04 PM
tbenne3
derivative question
Quote:

Originally Posted by skeeter
$\displaystyle \frac{d}{dx} \frac{x^3-2x}{\sqrt{2x+1}} = \frac{\sqrt{2x+1} \cdot (3x^2-2) - (x^3-2x) \cdot \frac{1}{\sqrt{2x+1}}}{2x+1}$

multiply by $\displaystyle \frac{\sqrt{2x+1}}{\sqrt{2x+1}}$ to clear the inner fraction ...

$\displaystyle \frac{(2x+1)(3x^2-2) - (x^3-2x)}{(2x+1)^{\frac{3}{2}}}$

$\displaystyle \frac{6x^3+3x^2-4x-2 - x^3 + 2x}{(2x+1)^{\frac{3}{2}}}$

$\displaystyle \frac{5x^3 + 3x^2 - 2x - 2}{(2x+1)^{\frac{3}{2}}}$

did he use the chain rule to come up with 1/sqrt(2x+1)
• Mar 4th 2010, 07:08 PM
tonio
Quote:

Originally Posted by tbenne3
did he use the chain rule to come up with 1/sqrt(2x+1)

If you mean in the first line's numerator the answer is no: that is simply the derivative of $\displaystyle \sqrt{2x+1}$

Tonio
• Mar 4th 2010, 07:10 PM
Yes tbenne3,

skeeter calculated

$\displaystyle \frac{d}{dx}\sqrt{2x+1}=\frac{d}{dx}\left(2x+1\rig ht)^\frac{1}{2}$

$\displaystyle =\frac{du}{dx}\frac{d}{du}u^{\frac{1}{2}}$
• Mar 4th 2010, 08:05 PM
tbenne3
no i was talking about on the very first line to the far right where it has 1/sqrt(2x+1)
• Mar 4th 2010, 09:04 PM
tbenne3
how did he get 1/sqrt(2x+1) in the very first step??????
• Mar 5th 2010, 01:44 AM
$\displaystyle \frac{d}{dx}\left(\frac{a(x)}{b(x)}\right)=\frac{b \frac{da}{dx}-a\frac{db}{dx}}{b^2}$
The $\displaystyle \frac{1}{\sqrt{2x+1}}$ is $\displaystyle \frac{db}{dx}$
as I described above by letting $\displaystyle u=2x+1$