# How do you find the derivative of xsqrt (1-x)?

Mar 11, 2018

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

#### Explanation:

The expression is product of two functions in $x$.

Denoting these by $f \left(x\right)$ and $g \left(x\right)$, respectively,

the first is
$f \left(x\right) = x$

and the second is
$g \left(x\right) = \sqrt{1 - x}$

$g \left(x\right)$ is a compound function (ie ie a function of a function)

The derivative of the expression is

$f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

The derivative of the first function is straightforward
$f ' \left(x\right) = 1$

The derivative of the second is trickier because it is a compound function. This requires the chain rule. The outer function is the square root function, and the inner function is the polynomial $\left(1 - x\right)$

writing the compound function as
$h \left(j \left(x\right)\right)$ ($h$ of $j$ of $x$), the derivative is

$h ' \left(j \left(x\right)\right) j ' \left(x\right)$

That is, the derivative of the outer function evaluated at the inner function times the derivative of the inner function

It makes things simpler to rewrite $g \left(x\right)$ using index notation, that is

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

Evaluating the outer function is the straightforward application of the rules of polynomial differentiation applied to its index, that is

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

And the derivative of the inner function is
$j ' \left(x\right) = - 1$

So the derivative of the compound function $g \left(x\right)$ is

$g ' \left(x\right) = h ' \left(j \left(x\right)\right) j ' \left(x\right) = \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}} \left(- 1\right)$
$= - \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}}$

Or, if you prefer, reverting to the square root notation and noting the negative index

$g ' \left(x\right) = - \frac{1}{2 \sqrt{1 - x}}$

So the overall derivative is

$f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

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

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

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

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

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

Mar 11, 2018

(2-3x)/(2sqrt(1-x)

#### Explanation:

$\textcolor{red}{\frac{d}{\mathrm{dx}} \left(u \cdot v\right) = u \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{du}}{\mathrm{dx}}}$
and,$\frac{d}{\mathrm{dx}} \left(\sqrt{1 - x}\right) = \frac{d}{\mathrm{dx}} {\left(1 - x\right)}^{\frac{1}{2}} = \frac{1}{2} \cdot {\left(1 - x\right)}^{- \frac{1}{2}} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - x}}$
$y = x \cdot \sqrt{1 - x}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{1 - x}\right) + \sqrt{1 - x} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \frac{1}{2 \sqrt{1 - x}} \left(- 1\right) + \sqrt{1 - x} \cdot 1$
=(-x)/(2sqrt(1-x))+sqrt(1-x)=(-x+2(1-x))/(2sqrt(1-x)
=(-x+2-2x)/(2sqrt(1-x))=(2-3x)/(2sqrt(1-x)