# How do you use the quotient rule to find the derivative of y=(1+sqrt(x))/(1-sqrt(x)) ?

Sep 2, 2014

$y ' = \frac{1}{\sqrt{x}} \cdot \frac{1}{{\left(1 - \sqrt{x}\right)}^{2}}$

Explanation :

Using Quotient Rule, which is

$y = f \frac{x}{g} \left(x\right)$, then

$y ' = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Similarly following for the given problem,

$y = \frac{1 + \sqrt{x}}{1 - \sqrt{x}}$

$y ' = \frac{\left(1 - \sqrt{x}\right) \left(\frac{1}{2 \sqrt{x}}\right) - \left(1 + \sqrt{x}\right) \left(- \frac{1}{2 \sqrt{x}}\right)}{{\left(1 - \sqrt{x}\right)}^{2}}$

$y ' = \frac{1}{2 \sqrt{x}} \cdot \frac{1 - \sqrt{x} + 1 + \sqrt{x}}{{\left(1 - \sqrt{x}\right)}^{2}}$

$y ' = \frac{1}{2 \sqrt{x}} \cdot \frac{2}{{\left(1 - \sqrt{x}\right)}^{2}}$

$y ' = \frac{1}{\sqrt{x}} \cdot \frac{1}{{\left(1 - \sqrt{x}\right)}^{2}}$