# How do you differentiate f(x)=x^2*sqrt(x-2) using the product rule?

Jan 19, 2016

$f ' \left(x\right) = \frac{5 {x}^{2} - 8 x}{2 \sqrt{x - 2}}$

#### Explanation:

using the 'product rule' and the 'chain rule ' :

rewrite f(x) = ${x}^{2} \sqrt{x - 2} = {x}^{2.} {\left(x - 2\right)}^{\frac{1}{2}}$

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

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

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

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

[ common factor of ${\left(x - 2\right)}^{- \frac{1}{2}}$ ]

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

# = (x - 2 )^(-1/2) [ 1/2 x^2 + 2x^2 - 4x ]

<$= {\left(x - 2\right)}^{- \frac{1}{2}} \left[\frac{1}{2} \left({x}^{2} + 4 {x}^{2} - 8 x\right)\right]$

$\Rightarrow f ' \left(x\right) = {\left(x - 2\right)}^{- \frac{1}{2}} \frac{1}{2} \left(5 {x}^{2} - 8 x\right) = \frac{5 {x}^{2} - 8 x}{2 \sqrt{x - 2}}$