# How do you differentiate (3x-2)/(2x+1)^(1/2)?

Jun 17, 2018

$\frac{3 x + 5}{\left(2 x + 1\right) \cdot \sqrt{2 x + 1}}$

#### Explanation:

By the Quotient rule we get

$\frac{3 \sqrt{2 x + 1} - \left(3 x - 1\right) \cdot \left(\frac{1}{2}\right) \cdot {\left(2 x + 1\right)}^{- \frac{1}{2}} \cdot 2}{2 x + 1}$
multiplying numerator and denominator by $\sqrt{2 x + 1}$

we get

$\frac{\left(3 \left(2 x + 1\right) - \left(3 x - 2\right)\right)}{\left(2 x + 1\right) \cdot \sqrt{2 x + 1}}$
simplifying we get

$\frac{3 x + 5}{\left(2 x + 1\right) \sqrt{2 x + 1}}$

Jun 17, 2018

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

#### Explanation:

With the quotient rule.

$f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$
$f ' \left(x\right) = \frac{u ' \left(x\right) v \left(x\right) - u \left(x\right) v ' \left(x\right)}{v {\left(x\right)}^{2}}$
$f \left(x\right) = \frac{3 x + 2}{{\left(2 x + 1\right)}^{\frac{1}{2}}}$
$u \left(x\right) = 3 x + 2 \mathmr{and} u ' \left(x\right) = 3$
$v \left(x\right) = {\left(2 x + 1\right)}^{\frac{1}{2}} \mathmr{and} v ' \left(x\right) = \frac{1}{2} \cdot 2 \cdot {\left(2 x + 1\right)}^{- \frac{1}{2}} = {\left(2 x + 1\right)}^{- \frac{1}{2}}$
$f ' \left(x\right) = \frac{3 \cdot {\left(2 x + 1\right)}^{\frac{1}{2}} - \left(3 x + 2\right) {\left(2 x + 1\right)}^{- \frac{1}{2}}}{2 x + 1}$
$f ' \left(x\right) = \frac{3 \cdot \left(2 x + 1\right) - \left(3 x + 2\right)}{2 x + 1} ^ \left(\frac{3}{2}\right)$
$f ' \left(x\right) = \frac{3 x + 1}{2 x + 1} ^ \left(\frac{3}{2}\right)$