# What is the derivative of f(x)=((x^2-3x)/(2x-3))*ln(x^2-3x)?

Dec 3, 2015

$f ' \left(x\right) = \frac{{\left(2 x - 3\right)}^{2} + \left(2 {x}^{2} - 6 x + 9\right) \cdot \ln \left({x}^{2} - 3 x\right)}{2 x - 3} ^ 2$

#### Explanation:

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

We can approach this by letting

$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$ $f ' \left(x\right) = g \left(x\right) \cdot h ' \left(x\right) + g ' \left(x\right) \cdot h \left(x\right) \text{ " " " " " " " " " } \left(1\right)$

color(red)(g(x) = ((2x^2 -3x)/(2x-3))" " " " " " " " (2)

color(blue)(h(x) = ln(x^2 -3x) " " " " " " " " "(3)

======================================

Let differentiate color(red)(g(x) = ((x^2-3x)/(2x-3)) using quotient rule

$g ' \left(x\right) = \frac{\left(2 x - 3\right) \left(2 x - 3\right) - \left(2 \left({x}^{2} - 3 x\right)\right)}{2 x - 3} ^ 2$
$g ' \left(x\right) = \frac{\left(4 {x}^{2} - 6 x - 6 x + 9\right) - \left(2 {x}^{2} - 6 x\right)}{2 x - 3} ^ 2$
$g ' \left(x\right) = \frac{4 {x}^{2} - 12 x + 9 - 2 {x}^{2} + 6 x}{2 x - 3} ^ 2$
color(red)(g'(x) = [2x^2-6x+9]/(2x-3)^2 " " " " " " " " " " " (4)

Let differentiate color(blue)(h(x) = ln(x^2 -3x)
Note: $f \left(x\right) = \ln \left(x\right) \Leftrightarrow f ' \left(x\right) = \frac{\mathrm{dx}}{x}$

color(blue)(h'(x) = (2x-3)/(x^2-3x) " " " " " " " " " " (5)

==============
Now we can put it together using $\left(1\right)$

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

$f ' \left(x\right) = \frac{{x}^{2} - 3 x}{2 x - 3} \cdot \textcolor{b l u e}{\frac{2 x - 3}{{x}^{2} - 3 x}} + \textcolor{red}{\frac{2 {x}^{2} - 6 x + 9}{2 x - 3} ^ 2} \cdot \ln \left({x}^{2} - 3 x\right)$
$f ' \left(x\right) = \cancel{\frac{\left({x}^{2} - 3 x\right)}{2 x - 3} \cdot \frac{\textcolor{b l u e}{\left(2 x - 3\right)}}{{x}^{2} - 3 x}} + \textcolor{red}{\frac{2 {x}^{2} - 6 x + 9}{2 x - 3} ^ 2} \cdot \ln \left({x}^{2} - 3 x\right)$
$f ' \left(x\right) = 1 + \left[\textcolor{red}{\frac{2 {x}^{2} - 6 x + 9}{2 x - 3} ^ 2} \cdot \ln \left({x}^{2} - 3 x\right)\right]$
$f ' \left(x\right) = \frac{{\left(2 x - 3\right)}^{2} + \left(2 {x}^{2} - 6 x + 9\right) \cdot \ln \left({x}^{2} - 3 x\right)}{2 x - 3} ^ 2$