# What is the derivative of xsqrt(1-x)?

Jun 2, 2018

$= \setminus \frac{2 - 3 x}{2 \setminus \sqrt{1 - x}}$

#### Explanation:

Question: $\setminus \frac{d}{\mathrm{dx}} x {\left(1 - x\right)}^{\setminus \frac{1}{2}}$

Use product rule, $\setminus \frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$, power rule $\setminus \frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ and chain rule, $\setminus \frac{d}{\mathrm{dx}} \left[f ' \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

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

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

$= \setminus \sqrt{1 - x} + x \cdot \left[\setminus \frac{- 1}{2} \cdot \setminus \frac{1}{\setminus \sqrt{1 - x}}\right]$

$= \setminus \sqrt{1 - x} + \setminus \frac{- x}{2 \setminus \sqrt{1 - x}}$

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

$= \setminus \frac{2 - 3 x}{2 \setminus \sqrt{1 - x}}$