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