# What is the derivative of (sqrtx-1)/sqrtx?

##### 2 Answers
Jan 22, 2016

$\frac{1}{2 \sqrt{{x}^{3}}}$

#### Explanation:

The quotient rule states that

$\frac{d}{\mathrm{dx}} \left[f \frac{x}{g} \left(x\right)\right] = \frac{g \left(x\right) \cdot f ' \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$

So application hereof in this particular case yields

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

$= \frac{\frac{1}{2} - \frac{1}{2} + \frac{1}{2 \sqrt{x}}}{x}$

$= \frac{1}{2 x \sqrt{x}}$

Jan 22, 2016

$\frac{1}{2 {x}^{\frac{3}{2}}}$

#### Explanation:

Begin by simplifying the function.

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

To differentiate from here, simply use the power rule.

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

This can also be written as

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