# What is the derivative of (3+2x)^(1/2)?

May 2, 2018

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

#### Explanation:

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left({\left(3 + 2 x\right)}^{\frac{1}{2}}\right)$

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

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

May 2, 2018

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

#### Explanation:

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

(apply the chain rule)

$u = 3 + 2 x$

$u ' = 2$

$f \left(u\right) = {u}^{\frac{1}{2}}$

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

Hence:

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

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

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