# How do you differentiate f(x) = (sqrtx)/(-x^2-2x+1) using the quotient rule?

Aug 15, 2016

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

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

Given $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} \text{ then}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$
$\textcolor{b l u e}{\text{------------------------------------------------------------}}$

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

and $h \left(x\right) = - {x}^{2} - 2 x + 1 \Rightarrow h ' \left(x\right) = - 2 x - 2$
$\textcolor{b l u e}{\text{------------------------------------------------------------}}$
substitute these values into (A)

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

simplifying the numerator.

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

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

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