differentiate f(x) = sqrt[ (3x+2)/ (2x-1)]

f(x)= sqrt[
(3x+2)/(2x-1)]

First we will separate the numerator and
denominator.

==> f(x) = sqrt(3x+2) /
sqrt(2x-1)

Now we will use the quotient rule to find the
derivative.

==> Let f(x) = u/v such
that:

==> u= sqrt(3x+2) ==> u' =
3/2sqrt(3x+2)

==> v= sqrt(2x-1) ==> v' =
1/sqrt(2x-1)

Now we know
that:

f'(x)= u'v - uv' /
v^2

==> f'(x) = [ 3sqrt(2x-1)/2sqrt(3x+2)  -
sqrt(3x+2)/sqrt(2x-1) / (2x-1)

==> f'(x) = (
3(2x-1) - 2(3x+2) / 2(2x-1)*sqrt(3x+2)(2x-1)

==>
f'(x) = ( 6x -3 - 6x -4) /
2(2x-1)*sqrt(3x+2)(2x-1)

==> f'(x)=
-7/2(2x-1)^3/2 *(3x+2)^1/2