# Thread: My 2 worst enemies combined...

1. ## 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

2. Use Quotient rule with

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

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

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

3. 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.

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

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

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

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

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

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

-Dan