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

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Answer 1 · 2016-01-19T16:33:27

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

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:

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

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