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

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We have to find the derivative of f(x) = sqrt[(3x+2)/(2x-1)]

Use the chain and the the quotient rules.

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

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

f'(x) = (1/2)*[1/sqrt [(3x+2)/(2x-1)]*(-7)/(2x - 1)^2]

f'(x) = (-7/2)/[sqrt [(3x+2)/(2x-1)]*(2x - 1)^2]

The required derivative is (-7/2)*1/[sqrt((3x+2)/(2x-1))*(2x - 1)^2]

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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