# How do you use the product Rule to find the derivative of x+1) (sqrt(2x-1))?

Aug 16, 2015

${y}^{'} = \frac{3 x}{\sqrt{2 x - 1}}$

#### Explanation:

Since your function can be written as the product of two other functions

$y = {\underbrace{\left(x + 1\right)}}_{\textcolor{red}{f \left(x\right)}} \cdot {\underbrace{{\left(2 x - 1\right)}^{\frac{1}{2}}}}_{\textcolor{g r e e n}{g \left(x\right)}}$

you can differentiate it by using the product rule

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}$

This means that you can write

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(x + 1\right)\right] \cdot {\left(2 x - 1\right)}^{\frac{1}{2}} + \left(x + 1\right) \cdot \frac{d}{\mathrm{dx}} {\left(2 x - 1\right)}^{\frac{1}{2}}$

You can calculate $\frac{d}{\mathrm{dx}} {\left(2 x - 1\right)}^{\frac{1}{2}}$ by using the chain rule for ${u}^{\frac{1}{2}}$, with $u = \left(2 x - 1\right)$.

$\frac{d}{\mathrm{dx}} \left({u}^{\frac{1}{2}}\right) = \frac{d}{\mathrm{du}} {u}^{\frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

d/dx(u^(1/2) = 1/2u^(-1/2) * d/dx(2x-1)

$\frac{d}{\mathrm{dx}} {\left(2 x - 1\right)}^{\frac{1}{2}} = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot {\left(2 x - 1\right)}^{- \frac{1}{2}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}$

Take this back to the calculation of your target derivative to get

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

This can be simplified to give

${y}^{'} = {\left(2 x - 1\right)}^{- \frac{1}{2}} \cdot \left(2 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} + x + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}}\right)$

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

This is equivalent to

${y}^{'} = \textcolor{g r e e n}{\frac{3 x}{\sqrt{2 x - 1}}}$