# How do you find the derivative of (2sqrtx - 1)/(2sqrtx)?

Jan 8, 2017

d/(dx) (frac (2sqrt(x)-1) (2sqrt(x)) )= 1/(4sqrt(x^3)

#### Explanation:

We could use the quotient rule, stating that:

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

It is however easier to write the function as:

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

so that:

d/(dx) (frac (2sqrt(x)-1) (2sqrt(x)) )= d/(dx) ( 1-1/2x^(-1/2)) = 1/4x^(-3/2) =1/(4sqrt(x^3)

Jan 8, 2017

Break it into two fractions, the first becomes the derivative of a constant, and the second can be differentiated as a negative power.

#### Explanation:

Break into two fractions:

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

Simplify:

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

The derivative of a constant is zero and use the power rule on the second term:

$\frac{d \left(\frac{2 \sqrt{x} - 1}{2 \sqrt{x}}\right)}{\mathrm{dx}} = \frac{1}{4} {x}^{- \frac{3}{2}}$