f(x) = sqrt [(2x+1) / (2x-1)]
find f'(x)
I'm using chain rule, but I'm not sure to bring the ^(1/2) to both top and bottom terms and then apply the quotiant rule.
any help wouldbe wonderful.
I find that the chain rule makes more sense when you use a u-substitution. Just set the 'inside' function equal to u:
$\displaystyle u=\frac{2x+1}{2x-1}$
$\displaystyle y=\sqrt{u}$
Since u is a function of x, the chain rule gives:
$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
$\displaystyle \frac{dy}{du}=\frac{1}{2\sqrt{u}}$
$\displaystyle \frac{du}{dx}=\frac{d}{dx}\frac{2x+1}{2x-1}$
Do you see how this works?
If you want to consider other simplifications,
you can try a few tricks such as..
$\displaystyle \sqrt{\frac{2x+1}{2x-1}}=\frac{\sqrt{2x+1}}{\sqrt{2x-1}}$
$\displaystyle =\frac{\sqrt{2x+1}\sqrt{2x-1}}{\sqrt{2x-1}\sqrt{2x-1}}$
$\displaystyle =\frac{\sqrt{(2x+1)(2x-1)}}{2x-1}=\frac{\sqrt{4x^2-1}}{2x-1}$
Then differentiate that.
Or...
$\displaystyle \frac{\sqrt{2x+1}}{\sqrt{2x-1}}=(2x+1)^{\frac{1}{2}}(2x-1)^{-\frac{1}{2}}$
Let $\displaystyle y=2x-1$
$\displaystyle \frac{dy}{dx}=2$
$\displaystyle \frac{d}{dx}[(y+2)^{\frac{1}{2}}y^{-\frac{1}{2}}]$
$\displaystyle =\frac{dy}{dx}\frac{d}{dy}[(y+2)^{\frac{1}{2}}y^{-\frac{1}{2}}]=2\frac{d}{dy}[(y+2)^{\frac{1}{2}}y^{-\frac{1}{2}}]$
Then use the "product rule".
[Math]\left(\sqrt{\frac{2x+1}{2x-1}}\right)'=\left(\frac{\sqrt{2x+1}}{\sqrt{2x-1}}\right)'
[/tex] $\displaystyle = \frac{\left(\sqrt{2x+1}\right)'\left(\sqrt{2x-1}\right)-\left(\sqrt{2x+1}\right)\left(\sqrt{2x-1}\right)'}{\left(\sqrt{2x-1}\right)^2}$ $\displaystyle = \dfrac{\frac{\sqrt{2x-1}}{\sqrt{2x+1}}-\frac{\sqrt{2x+1}}{\sqrt{2x-1}}}{2x-1}$ $\displaystyle = \frac{\sqrt{2x-1}\sqrt{2x-1}-\sqrt{2x+1}{\sqrt{2x+1}}}{(2x-1)\sqrt{2x+1}\sqrt{2x-1}}$ $\displaystyle = \dfrac{(2x-1)-(2x+1)}{\sqrt{(2x-1)^3}\sqrt{2x+1}}$ $\displaystyle = \boxed{\dfrac{-2}{\sqrt{(2x-1)^3}\sqrt{2x+1}}}$.