# How do you differentiate y = (x+1) (sqrt (2x-1))?

Feb 20, 2016

The simplified condensed solution is:
$\sqrt{2 x - 1} + \frac{x + 1}{\sqrt{2 x - 1}} = \textcolor{red}{\frac{3 x}{\sqrt{2 x - 1}}}$

#### Explanation:

Let $h \left(x\right) = f \left(x\right) g \left(x\right)$ where $f \left(x\right) = \left(x + 1\right) \text{ and } g \left(x\right) = \sqrt{2 x - 1}$

Now by the product rule we have:

$\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} = h ' \left(x\right) = \textcolor{b l u e}{f ' \left(x\right)} g \left(x\right) + \textcolor{g r e e n}{g ' \left(x\right)} f \left(x\right)$
$\textcolor{b l u e}{f ' \left(x\right)} = \textcolor{b l u e}{1}$
$\textcolor{g r e e n}{g ' \left(x\right)} = \textcolor{g r e e n}{\left(\frac{1}{2 \sqrt{2 x - 1}}\right) \left(\frac{d \left(2 x - 1\right)}{\mathrm{dx}}\right) \left(x + 1\right)}$
color(green)(g'(x) =((x+1)/(sqrt(2x-1)))

$h ' \left(x\right) = \sqrt{2 x - 1} + \left(\frac{1}{2 \sqrt{2 x - 1}}\right) \left(\frac{d \left(2 x - 1\right)}{\mathrm{dx}}\right) \left(x + 1\right)$
$h ' \left(x\right) = \sqrt{2 x - 1} + \left(\frac{x + 1}{\sqrt{2 x - 1}}\right)$ you can further simplify
$h ' \left(x\right) = \textcolor{red}{\frac{3 x}{\sqrt{2 x - 1}}}$