Asked by mike

use logarithmic differentiation to find dy/dx for: [(x^2)(e^2)(x)] / [3√(2x-5)]
Answered by Steve
I assume you mean

y = [x^2 e^(2x)]/ [3√(2x-5)]

ln y = 2lnx + 2x - 3/2 ln(2x-5)
1/y y' = 2/x + 2 - 3/(2x-5)
1/y y' = (4x^2-9x-10)/[x(2x-5)]

y' = [x^2 e^(2x)]/ [3√(2x-5)] * (4x^2-9x-10)/[x(2x-5)]

x e^(2x) (4x^2-7x-10)
----------------------------
      3(2x-5)^(3/2)
Answered by Steve
oops. That should have been

ln y = 2lnx + 2x - ln3 - 1/2 ln(2x-5)

and also 4x^2-7x-10 in the numerator
Answered by mike
how did you get 4x^2-7x-10?
Answered by Steve
a little algebra. You didn't get that? Show whatcha got.
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