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

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- Mar 4th 2010, 07:04 PMtbenne3derivative question
- Mar 4th 2010, 07:08 PMtonio
- Mar 4th 2010, 07:10 PMArchie Meade
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 PMtbenne3
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 PMtbenne3
**how did he get 1/sqrt(2x+1) in the very first step??????** - Mar 5th 2010, 01:44 AMArchie Meade
The Quotient Rule is

$\displaystyle \frac{d}{dx}\left(\frac{a(x)}{b(x)}\right)=\frac{b \frac{da}{dx}-a\frac{db}{dx}}{b^2}$

The $\displaystyle \frac{1}{\sqrt{2x+1}}$ is $\displaystyle \frac{db}{dx}$

as I described above by letting $\displaystyle u=2x+1$