Question

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

Answer 1

`f'(x) = (15)/((3x+3)(2x-8))`

Explanation:

Let's do this. :)

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

`sqrt(x) = x^(1/2)`

and the following logarithmic rule:

`ln(a^r) = r * ln(a)`

Thus, we can simplify as follows:

`ln sqrt((2x-8)/(3x+3)) = ln [((2x-8)/(3x+3)) ^(1/2)] = 1/2 * ln ((2x-8)/(3x+3))`

This is easier already.

As next, we will need the chain rule:

`f(x) = 1/2 ln u " "` where `" " u = (2x-8)/(3x+3)`

Thus, the derivative of `f(x)` is the derivative of `1/2 ln u` multiplied with the derivative of `u`.

Let's compute both:

`[1/2 ln u]' = 1/2 * 1 /u = 1 / 2 * 1 / ((2x-8)/(3x+3)) = 1 / 2 * (3x+3)/(2x-8)`

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

for `u = g/h`, the derivative is `u' = (g' h - h' g) / h^2`

In this case, we have

`u' = (2(3x+ 3 ) - 3(2x - 8)) / ((3x+3)^2) = 30 / (3x+3)^2`

In total, we have:

`f'(x) = [1/2 ln u]' * u' = 1 / 2 * (3x+3)/(2x-8) * (30) / (3x+3)^2 = (15)/((3x+3)(2x-8))`