# How do you find the derivative for sqrt(4x)/(x-1)?

May 14, 2015

Quocient rule time!

Let's just rewrite the two functions the compose your fraction: the first one is the same as $\sqrt{2 \cdot 2 \cdot x}$, which is the same as $2 \sqrt{x}$.

The quocient rule determines that:

If $y = \frac{f \left(x\right)}{g \left(x\right)}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{f ' \left(x\right) \cdot g \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{f {\left(x\right)}^{2}}$.

Now, let's just proceed according to the quocient rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1}{{\left(x - 1\right)}^{2}} \cdot \left(x - 1\right) - 2 \sqrt{x} \cdot 1}{4 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{x - 1}{x - 1} ^ 2 - 2 \sqrt{x}}{4 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1}{x - 1} - 2 \sqrt{x}}{4 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 x}{x - 1} - \frac{\sqrt{x}}{2 x}$

Finally:

(dy)/(dx) = (4x)/(x-1)-1/(2sqrt(x)