Logarithmic Function with Sqare Root?

Question

(x)=ln(sqroot((2x-3)/(3x+6))

f'(x)=

I believe you take the 1/2 on the outside but the computation after that is screwing me up! Any help would be great!

Sorry the problem is f(x)=ln(sqroot((2x-3)/(3x+6))

forgot the f(x)

Answer 1

f(x) = ln{[(2x-3)/(3x+6)]^(1/2)} = (1/2)ln[(2x-3)/(3x+6)]

= (1/2)ln(2x-3) - (1/2)ln(3x+6)

f'(x) = (1/2)[2/(2x-3)] - (1/2)[3/(3x+6)]
f'(x) = (1/2)[(2/(2x-3)) - (3/(3x+6))]
f'(x) = (1/2)[(2(3x+6) - 3(2x-3))/(2x-3)(3x+6)]
f'(x) = (1/2)[21/(2x-3)(3x+6)]
f'(x) = 21/(2(2x-3)(3x+6))
f'(x) = 7/(2(2x-3)(x+2))
or f'(x) = 7/(4x² +2x - 12)

Answer 2

The function can be rewritten as:

(1/2)ln(2x-3) - (1/2)ln(3x+6) because of logarithmic rules.

Then, the derivative of (1/2)ln(2x-3) is equal to
(1/2)*(1/(2x-3))*2 which reduces to (1/(2x-3))

The derivative of (1/2)ln(3x+6) equals
(1/2)*(1/(3x+6))*3 which reduces to (1/(2x+4))

(1/(2x-3)) - (1/(2x+4)) is the derivative

[Can be written as (2x-3)^(-1) - (2x+4)^(-1)]

Answer 3

f(x) = ln(((2x-3)/(3x+6))^(1/2)
f(x) = (1/2)ln(((2x-3)/(3x+6)))
f(x) = (1/2)(ln(2x-3) - ln(3x+6))
f'(x) = (1/2)(2 / (2x - 3) - 3 / (3x + 6) )
f'(x) = (1/2)(2(3x + 6) - 3(2x - 3) / (2x - 3)(3x + 6) )
f'(x) = (1/2)(6x + 12 - 6x + 9) / (2x - 3)(3x + 6)
f'(x) = (1/2)21/ (2x - 3)(3x + 6)

Answer 4

sqrt(x)=x^.5 so ln[sqrt(x)] = ln(x^.5) = .5ln(x)

So ln[sqrt((2x-3)/(3x+6)] = ln[((2x-3)/(3x+6))^.5]

ln[((2x-3)/(3x+6))^.5] = .5ln[(2x-3)/(3x+6)] = .5[ln(2x-3) - ln(3x+6)]

Answer 5

you need to do the product formulation. see it as x*sqrt(x-3). then do... (first)(der. of 2nd) + (2nd)(der. of first). then multiply jointly or ingredient so remedy for x. wish this facilitates. that is a sprint complicated. sorry. :/

Answer 6

Sorry but this is not a question just a statement.

What do you want to do with f(x)?