Finding the derivative: 3x-2/(2x+1)^1/2?

Question

All that is needed is finding the derivative and simplifying the answer. After attempting it I couldn't come up with a sensible answer. 3x-2/(2x+1)^1/2. (The denominator is the square root of 2x+1).

Answer 1

Since it's difficult to tell, I'm gonna assume the 3x and -2 are together. So your question is then

(3x-2)/(2x+1)^(1/2)

Let f(x) = 3x - 2 and g(x) = (2x+1)^(1/2)

The formula for the derivative of a fraction is, verbally:
"The derivative of the top times the bottom minus the derivative of the bottom times the top, over the bottom squared."

f'(x) = 3 and
g'(x) = [1 / 2sqrt(2x+1)] * 2 = 1/sqrt(2x+1)
(note: we multiply 2 for g'(x) because of the chain rule.

Then, we apply our formula. The derivative is:

{ [3*(2x+1)^(1/2)] - [ 1/sqrt(2x+1) * (3x-2)] } / (2x+1)

Note: the bottom is (2x+1) because squaring a square root becomes itself. I'm taking sqrt(2x+1) squared.

Answer 2

Start by rewriting in non fractional form

(3x-2)(2x+1)^(-1/2)

using "d" for the derivative symbol

duv = (du)(v) + (dv)(u)
d(3x-2) = 3
d((2x+1)^(-1/2) = -1/2[(2x+1)^(3/2)](2)

which leads us to
(3)/[(2x+1)^(-1/2)] + [-(3x-2)/[(2x+1)^(3/2)]

This site seriously needs some way to enter math notation - this is too obnoxious!

Answer 3

this is what I got.

3/(sqrt(2x+1))-(3x-2)/(2x+1)^(3/2)

Answer 4

i would rewrite it like this
(3x-2)*(2x+1)^(-1/2)
then work it
3*((2x+1)^(-1/2)) + (3x-2)(-1/2(2x+1)2)
and simpifly as you like
3*((2x+1)^(-1/2)) + (3x-2)(-2x-1)
3((2x+1)^(-1/2)) -6x^2 + x + 2

Answer 5

You haven't said what the equation should equal? Or am I missing something?

Answer 6

ahh how i love calculus =D i love derivatives..integrals are icky

Answer 7

Us the QUOTIENT RULE:

f(x)/g(x) =

((g(x)*f'(x))-(g'(x)*f(x)))/(g(x))^2