# How do you differentiate f(x)= (x^2+3x-6)/(2x-4) twice using the quotient rule?

Dec 15, 2015

Quotient Rule:
$\frac{d}{\mathrm{dx}} \left\{f \frac{x}{g} \left(x\right)\right\} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$

Original Equation:
$\frac{{x}^{2} + 3 x - 6}{2 x - 4}$

First, find the derivative of the numerator:
$2 x + 3$

Times that by the denominator:
$\left(2 x + 3\right) \left(2 x - 4\right)$
$4 {x}^{2} - 2 x - 12$
Find the derivative of the denominator:
$2$

Times it by the numerator:
$2 \left({x}^{2} + 3 x - 6\right)$
$2 {x}^{2} + 6 x - 12$

Subtract the two:
$4 {x}^{2} - 2 x - 12 - 2 {x}^{2} + 6 x - 12$
$2 {x}^{2} + 4 x - 24$

Square the denominator and divide by it:
$\frac{2 {x}^{2} + 4 x - 24}{2 x - 4} ^ 2$

You can't simplify, so
$f ' \left(x\right) = \frac{2 {x}^{2} + 4 x - 24}{2 x - 4} ^ 2$

Now repeat the above steps, but with $f ' \left(x\right)$:
$\frac{\left(4 x + 4\right) {\left(2 x - 4\right)}^{2} - \left(2 {x}^{2} + 4 x - 24\right) \left(2\right) \left(2 x - 4\right)}{2 x - 4} ^ 4$
$\frac{\left(4 x + 4\right) \left(2 x - 4\right) - \left(2 {x}^{2} + 4 x - 24\right) \left(2\right)}{2 x - 4} ^ 3$
$\frac{8 {x}^{2} - 8 x - 16 - 4 {x}^{2} - 8 x + 48}{2 x - 4} ^ 3$
$\frac{4 {x}^{2} - 16 x + 32}{2 x - 4} ^ 3$

$f ' ' \left(x\right) = \frac{4 {x}^{2} - 16 x + 32}{2 x - 4} ^ 3$