# What is the derivative of f(x) = ln sqrt ((2x-8 )/( 3x+3))?

Jan 19, 2016

$f ' \left(x\right) = \frac{15}{\left(3 x + 3\right) \left(2 x - 8\right)}$

#### Explanation:

Let's do this. :)

First of all, let me simplify the expression using the following power rule:

$\sqrt{x} = {x}^{\frac{1}{2}}$

and the following logarithmic rule:

$\ln \left({a}^{r}\right) = r \cdot \ln \left(a\right)$

Thus, we can simplify as follows:

$\ln \sqrt{\frac{2 x - 8}{3 x + 3}} = \ln \left[{\left(\frac{2 x - 8}{3 x + 3}\right)}^{\frac{1}{2}}\right] = \frac{1}{2} \cdot \ln \left(\frac{2 x - 8}{3 x + 3}\right)$

As next, we will need the chain rule:

$f \left(x\right) = \frac{1}{2} \ln u \text{ }$ where $\text{ } u = \frac{2 x - 8}{3 x + 3}$

Thus, the derivative of $f \left(x\right)$ is the derivative of $\frac{1}{2} \ln u$ multiplied with the derivative of $u$.

Let's compute both:

$\left[\frac{1}{2} \ln u\right] ' = \frac{1}{2} \cdot \frac{1}{u} = \frac{1}{2} \cdot \frac{1}{\frac{2 x - 8}{3 x + 3}} = \frac{1}{2} \cdot \frac{3 x + 3}{2 x - 8}$

To compute the derivative of $u$, let's use the quotient rule:

for $u = \frac{g}{h}$, the derivative is $u ' = \frac{g ' h - h ' g}{h} ^ 2$

In this case, we have

$u ' = \frac{2 \left(3 x + 3\right) - 3 \left(2 x - 8\right)}{{\left(3 x + 3\right)}^{2}} = \frac{30}{3 x + 3} ^ 2$

In total, we have:

$f ' \left(x\right) = \left[\frac{1}{2} \ln u\right] ' \cdot u ' = \frac{1}{2} \cdot \frac{3 x + 3}{2 x - 8} \cdot \frac{30}{3 x + 3} ^ 2 = \frac{15}{\left(3 x + 3\right) \left(2 x - 8\right)}$