# How do you find the derivative of G(x) = sqrtx (x^2 – x)^3?

Nov 8, 2017

Use chain rule and product rule. $= {\left({x}^{2} - x\right)}^{2} \left(\frac{{x}^{2} - x}{2 \sqrt{x}} + \left(6 x - 3\right) \sqrt{x}\right)$

#### Explanation:

As written, we use both the product rule and chain rule.

product rule: $f \left(x\right) = g \left(x\right) \cdot h \left(x\right) \to f ' = g ' \cdot h + g \cdot h '$
chain rule: $f \left(x\right) = g \left(h \left(x\right)\right) \to \frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{dh}}{\mathrm{dx}} \cdot \frac{\mathrm{dg}}{\mathrm{dh}}$

Then with $G \left(x\right) = \sqrt{x} {\left({x}^{2} - x\right)}^{3.} . .$

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

$= {\left({x}^{2} - x\right)}^{2} \left(\frac{{x}^{2} - x}{2 \sqrt{x}} + \left(6 x - 3\right) \sqrt{x}\right)$

Nov 8, 2017

$G ' \left(x\right) = \frac{1}{2} \sqrt{x} {\left({x}^{2} - x\right)}^{2} \left(13 x - 7\right)$

#### Explanation:

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

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

$G ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \text{product rule}$

$g \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}} \Rightarrow g ' \left(x\right) = \frac{1}{2} {x}^{- \frac{1}{2}} \leftarrow \text{chain rule}$

$h \left(x\right) = {\left({x}^{2} - x\right)}^{3} \leftarrow \text{differentiate using chain rule}$

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

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

$\textcolor{w h i t e}{\Rightarrow G} = \frac{1}{2} {x}^{- \frac{1}{2}} {\left({x}^{2} - x\right)}^{2} \left[6 x \left(2 x - 1\right) + {x}^{2} - x\right]$

$\textcolor{w h i t e}{\times \times} = \frac{1}{2} {x}^{- \frac{1}{2}} {\left({x}^{2} - x\right)}^{2} \left(13 {x}^{2} - 7 x\right)$

$\textcolor{w h i t e}{\times \times} = \frac{1}{2} \sqrt{x} {\left({x}^{2} - x\right)}^{2} \left(13 x - 7\right)$