# How do you find the derivative of  f(x)= (x+1)/sqrtx?

Jul 7, 2016

$f ' \left(x\right) = \frac{1}{2} \left({x}^{- \frac{1}{2}} - {x}^{- \frac{3}{2}}\right) ,$

OR, $f ' \left(x\right) = \frac{1}{2} \left\{\frac{1}{\sqrt{x}} - \frac{1}{x \cdot \sqrt{x}}\right\} = \frac{1}{2} \left\{\frac{x - 1}{x \sqrt{x}}\right\} = \frac{x - 1}{2 x \sqrt{x}} .$

#### Explanation:

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

Therefore, $f ' \left(x\right) = \frac{1}{2} \cdot {x}^{\frac{1}{2} - 1} + \left(- \frac{1}{2}\right) \cdot {x}^{- \frac{1}{2} - 1}$
$\therefore f ' \left(x\right) = \frac{1}{2} \left({x}^{- \frac{1}{2}} - {x}^{- \frac{3}{2}}\right) ,$

OR, $f ' \left(x\right) = \frac{1}{2} \left\{\frac{1}{\sqrt{x}} - \frac{1}{x \cdot \sqrt{x}}\right\} = \frac{1}{2} \left\{\frac{x - 1}{x \sqrt{x}}\right\} = \frac{x - 1}{2 x \sqrt{x}} .$