derivative question involving log?

Question

hi

i am asking this question again as i got little response last time

i would like to find out how to solve this problem.

i would like to find the derivative f ' (x) given f(x) = ln (9x-4/3x-5)

thanks for any help. it is much appreciated

Answer 1

f(x) = ln( (9x - 4)/(3x - 5) )

There are one of several ways to solve this, but I think you should solve it by first using log identities to manipulate it, and then differentiate after.

The log of a quotient is equal to the difference of logs, so

f(x) = ln(9x - 4) - ln(3x - 5)

Now, we differentiate, keeping in mind that d/dx ln(x) = 1/x, and that we have to use the chain rule.

f'(x) = 1/[9x - 4] {9} - 1/[3x - 5] {3}
f'(x) = 9/[9x - 4] - 3/[3x - 5]

If we wanted to create one fraction out of this (which is required for graph sketching), we give the fractions a common denominator and solve.

f'(x) = [ 9(3x - 5) - 3(9x - 4) ] / [(9x - 4)(3x - 5)]
f'(x) = [27x - 45 - 27x + 12] / [(9x - 4)(3x - 5)]
f'(x) = [-33] / [(9x - 4)(3x - 5)]

*edited for correct response*

Answer 2

Let u = (9x - 4) / (3x - 5)
du/dx = ((3x - 5).9 - 3.(9x - 4)) / (3x - 5)²
du/dx = (27x - 45 - 27x + 12) / (3x - 5)²
du/dx = -33 / (3x - 5)²
y = ln u
dy/du = 1 / u = (3x - 5) / (9x - 4)
dy/dx = (dy / du).(du / dx)
dy/dx = - 33.(3x - 5) / (9x - 4).(3x - 5)²
dy/dx = - 33 / [ (9x - 4).(3x - 5) ]

Answer 3

f(x) = ln( (9x-4)/(3x-5))

Remember
if y=u/v then y'=(u'v-v'u)/v^2


y'=(9(3x-5)-3(9x-4))/(3x-5)^2 * (3x-5)/(9x-4)


f'(x)=-33 / ((9x-4)(3x-5))


Hint: y=ln u
y'=u'/u