# How do you differentiate (x^2 + x + 3 )/ sqrt(x-3) using the quotient rule?

Nov 27, 2015

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

#### Explanation:

The quotient rule; given $f \left(x\right) \ne 0$

if $h \left(x\right) = f \frac{x}{g} \left(x\right)$ ; then $h ' \left(x\right) = \frac{g \left(x\right) \cdot f ' \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$

given $h \left(x\right) = \frac{{x}^{2} + x + 3}{\sqrt[]{x - 3}}$

let $f \left(x\right) = {x}^{2} + x + 3$
$\textcolor{red}{f ' \left(x\right) = 2 x + 1}$

let $g \left(x\right) = \sqrt[]{x - 3} = {\left(x - 3\right)}^{\frac{1}{2}}$

color(blue)(g'(x)= 1/2 (x-3)^(1/2-1)= 1/2 (x-3)^(-1/2)

h'(x)= [(x-3)^(1/2)*color(red)((2x+1))-color(blue)(1/2(x-3)^(-1/2))(x^2 +x+3)]/(root()[(x-3)]^2

Factor out the greatest common factor $\frac{1}{2} {\left(x - 3\right)}^{- \frac{1}{2}}$

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

$\implies h ' \left(x\right) = \frac{1}{2} \frac{\left({x}^{2} + x - 6 x - 3 - {x}^{2} - x - 3\right)}{x - 3} ^ \left(\frac{3}{2}\right)$
$h ' \left(x\right) = \frac{- 6 x - 6}{2 {\left(x - 3\right)}^{\frac{3}{2}}}$

$h ' \left(x\right) = - \frac{6 \left(x + 1\right)}{2 {\left(x - 3\right)}^{\frac{3}{2}}}$

$\textcolor{red}{h ' \left(x\right) = - \frac{3 \left(x + 1\right)}{{\left(x - 3\right)}^{\frac{3}{2}}}}$ Answer