My 2 worst enemies combined...

• Aug 27th 2006, 11:22 AM
becky
My 2 worst enemies combined...
Fractions and Square Roots...

Differentiate and simplify.

G(x) = square root of (3x+1/2x-1)

First, do I deal with this as a whole or break it down as two seperate functions: (3x+1) and (2x-1^1/2)
Second, I am not sure if I have written it to the correct power: 3x+1/2x-1^-1/2

-1/2(3x+1/2x-1)^-3/2
• Aug 27th 2006, 11:59 AM
Glaysher
Use Quotient rule with

$u = \sqrt{3x +1} = (3x + 1)^{\frac{1}{2}}$

$v = \sqrt{2x - 1} = (2x - 1)^{\frac{1}{2}}$

EDIT

My approach is slightly different from topsquark. You will need to use the chain rule twice with mine but they will both work in exactly the same way
• Aug 27th 2006, 12:02 PM
topsquark
Quote:

Originally Posted by becky
Fractions and Square Roots...

Differentiate and simplify.

G(x) = square root of (3x+1/2x-1)

First, do I deal with this as a whole or break it down as two seperate functions: (3x+1) and (2x-1^1/2)
Second, I am not sure if I have written it to the correct power: 3x+1/2x-1^-1/2

-1/2(3x+1/2x-1)^-3/2

You need the chain rule and quotient rule.

$G(u) = \sqrt{u}$ and $u(x) = \frac{3x+1}{2x-1}$.

$\frac{dG}{dx} = \frac{dG}{du} \cdot \frac{du}{dx}$.

So
$\frac{dG}{du} = \frac{d}{du}u^{1/2} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} = \frac{ \sqrt{2x-1}}{2 \sqrt{3x+1}}$

$\frac{du}{dx} = \frac{(3)(2x-1) - (3x+1)(2)}{(2x-1)^2} = \frac{-5}{(2x-1)^2}$

Finally: $\frac{dG}{dx} = \frac{ \sqrt{2x-1}}{2 \sqrt{3x+1}}\frac{-5}{(2x-1)^2}$

$\frac{dG}{dx} = \frac{-5}{2 \sqrt{3x+1}(2x-1)^{3/2}}$

-Dan