I'm going to assume you have the top and bottom grouped all together, so that it's
√((2x^2 - 1) / (3x^2 + 2))
Taking the first derivative gives
(1/2)((2x^2 - 1) / (3x^2 + 2))^(-1/2), times the derivative of the inside.
The inside is (2x^2 - 1) * (3x^2 + 2)^(-1), so take the first times the derivative of the second, plus the second times the derivative of the first. This gives you
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well... it's [2x^2 - 1 / 3x^2 +2]^.5
so chain rule the .5 and now you have:
.5[2x^2 -1 / 3x^2 +2][<<-- the derivitave of that]^(-.5)
I'm going to assume you have the top and bottom grouped all together, so that it's
√((2x^2 - 1) / (3x^2 + 2))
Taking the first derivative gives
(1/2)((2x^2 - 1) / (3x^2 + 2))^(-1/2), times the derivative of the inside.
The inside is (2x^2 - 1) * (3x^2 + 2)^(-1), so take the first times the derivative of the second, plus the second times the derivative of the first. This gives you
(2x^2 - 1) * -(3x^2 + 2)^-2 * 6x + (4x)(3x^2+2))^-1
which is
-6x(2x^2 - 1)(3x^2 + 2)^-2 + (4x)(3x^2+2))^-1
So combining it all, the derivative is:
(1/2)((2x^2 - 1) / (3x^2 + 2))^(-1/2) * [ -6x(2x^2 - 1)(3x^2 + 2)^-2 + (4x)(3x^2+2))^-1 ]
You can probably simplifiy it a little from there.
It would be easier to put it in this form
(2x^2) - 1/3x^2 + 2) ^ 1/2
= 1/2 * (2x^2) - 1/3x^2 + 2) ^ (-1/2) * d/dx(2x^2) - 1/3x^2 + 2)
= 1/2 * (2x^2) - 1/3x^2 + 2) ^ (-1/2) * (4x - 2/3)
Since they're different powers, I don't think you can reduce more than this.