# How do you find the derivative of the function  (3x-2)/(2x+1)^(1/2)?

Jul 26, 2017

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

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \text{ quotient rule}$

$g \left(x\right) = 3 x - 2 \Rightarrow g ' \left(x\right) = 3$

$\text{differentiate "(2x+1)^(1/2)" using the "color(blue)"chain rule}$

$h \left(x\right) = {\left(2 x + 1\right)}^{\frac{1}{2}} \Rightarrow h ' \left(x\right) = \frac{1}{2} {\left(2 x + 1\right)}^{- \frac{1}{2}} \times 2$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times} = {\left(2 x + 1\right)}^{- \frac{1}{2}}$

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

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{{\left(2 x + 1\right)}^{- \frac{1}{2}} \left(3 \left(2 x + 1\right) - \left(3 x - 2\right)\right)}{2 x + 1}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{3 x + 5}{2 x + 1} ^ \left(\frac{3}{2}\right)$