# How do you find the first and second derivatives of f(x)=-2/(x^2-2x)+(x+2)/(x-3) using the quotient rule?

Jan 19, 2018

$f ' \left(x\right) = \frac{4 x - 4}{{x}^{2} - 2 x} ^ 2 - \frac{5}{x - 3} ^ 2$
$f ' ' \left(x\right) = \frac{4 \left({x}^{2} - 2 x\right) - 4 {\left(2 x - 2\right)}^{2}}{{x}^{2} - 2 x} ^ 3 + \frac{10}{x - 3} ^ 3$

#### Explanation:

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

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[- \frac{2}{{x}^{2} - 2 x} + \frac{x + 2}{x - 3}\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{d}{\mathrm{dx}} \left[- \frac{2}{{x}^{2} - 2 x}\right] + \frac{d}{\mathrm{dx}} \left[\frac{x + 2}{x - 3}\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{\frac{d}{\mathrm{dx}} \left[2\right] \left({x}^{2} - 2 x\right) - 2 \frac{d}{\mathrm{dx}} \left[{x}^{2} - 2 x\right]}{{x}^{2} - 2 x} ^ 2 + \frac{\left(x - 3\right) \frac{d}{\mathrm{dx}} \left[x + 2\right] - \left(x + 2\right) \frac{d}{\mathrm{dx}} \left[x - 3\right]}{x - 3} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{0 \left({x}^{2} - 2 x\right) - 2 \left(2 x - 2\right)}{{x}^{2} - 2 x} ^ 2 + \frac{1 \left(x - 3\right) - 1 \left(x + 2\right)}{x - 3} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{- 2 \left(2 x - 2\right)}{{x}^{2} - 2 x} ^ 2 + \frac{x - 3 - x - 2}{x - 3} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{4 x - 4}{{x}^{2} - 2 x} ^ 2 + \frac{- 5}{x - 3} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{4 x - 4}{{x}^{2} - 2 x} ^ 2 - \frac{5}{x - 3} ^ 2$

$f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\frac{4 x - 4}{{x}^{2} - 2 x} ^ 2 - \frac{5}{x - 3} ^ 2\right]$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{d}{\mathrm{dx}} \left[\frac{4 x - 4}{{x}^{2} - 2 x} ^ 2\right] - \frac{d}{\mathrm{dx}} \left[\frac{5}{x - 3} ^ 2\right]$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{\frac{d}{\mathrm{dx}} \left[4 x - 4\right] {\left({x}^{2} - 2 x\right)}^{2} - \left(4 x - 4\right) \frac{d}{\mathrm{dx}} \left[{\left({x}^{2} - 2 x\right)}^{2}\right]}{{\left({x}^{2} - 2 x\right)}^{2}} ^ 2 - \frac{\frac{d}{\mathrm{dx}} \left[5\right] {\left(x - 3\right)}^{2} - 5 \frac{d}{\mathrm{dx}} \left[{\left(x - 3\right)}^{2}\right]}{{\left(x - 3\right)}^{2}} ^ 2$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{4 {\left({x}^{2} - 2 x\right)}^{2} - \left(4 x - 4\right) \frac{d}{\mathrm{dx}} \left[{\left({x}^{2} - 2 x\right)}^{2}\right]}{{x}^{2} - 2 x} ^ 4 - \frac{0 {\left(x - 3\right)}^{2} - 5 \frac{d}{\mathrm{dx}} \left[{\left(x - 3\right)}^{2}\right]}{x - 3} ^ 4$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{4 {\left({x}^{2} - 2 x\right)}^{2} - \left(4 x - 4\right) \left(2 \cdot {\left({x}^{2} - 2 x\right)}^{2 - 1} \frac{d}{\mathrm{dx}} \left[{x}^{2} - 2 x\right]\right)}{{x}^{2} - 2 x} ^ 4 - \frac{0 {\left(x - 3\right)}^{2} - 5 \left(2 \cdot {\left(x - 3\right)}^{2 - 1} \frac{d}{\mathrm{dx}} \left[x - 3\right]\right)}{x - 3} ^ 4$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{4 {\left({x}^{2} - 2 x\right)}^{2} - \left(4 x - 4\right) \left(2 \left({x}^{2} - 2 x\right) \left(2 x - 2\right)\right)}{{x}^{2} - 2 x} ^ 4 - \frac{- 5 \left(2 \left(x - 3\right) \left(1\right)\right)}{x - 3} ^ 4$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{4 {\left({x}^{2} - 2 x\right)}^{2} - 2 \left(4 x - 4\right) \left({x}^{2} - 2 x\right) \left(2 x - 2\right)}{{x}^{2} - 2 x} ^ 4 + \frac{10 \left(x - 3\right)}{x - 3} ^ 4$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{4 {\left({x}^{2} - 2 x\right)}^{2} - 4 \left(2 x - 2\right) \left({x}^{2} - 2 x\right) \left(2 x - 2\right)}{{x}^{2} - 2 x} ^ 4 + \frac{10}{x - 3} ^ 3$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{4 {\left({x}^{2} - 2 x\right)}^{2} - 4 {\left(2 x - 2\right)}^{2} \left({x}^{2} - 2 x\right)}{{x}^{2} - 2 x} ^ 4 + \frac{10}{x - 3} ^ 3$

$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{4 \left({x}^{2} - 2 x\right) - 4 {\left(2 x - 2\right)}^{2}}{{x}^{2} - 2 x} ^ 3 + \frac{10}{x - 3} ^ 3$