differential calculus
2007-12-08 23:08:38 · 6 answers · asked by citizen5 1 in Science & Mathematics ➔ Mathematics
d/dx ((5x+4)/(3x-2))
= {(3x-2)*5 - (5x+4)*3}/(3x-2)^2
= -22/(3x-2)^2 ...Final answer
2007-12-08 23:16:42 · answer #1 · answered by Nterprize 3 · 0⤊ 0⤋
do it by product rule (the guy at the top seems a little confused)
I remember it by "low dee high take high dee low all over the square of what's below" (it rhymes as well!)
that is d/dx of u/v = [v du - u dv] / v²
Therefore d/dx (5x + 4 / 3x - 2) =
[5(3x - 2) - 3(5x + 4)] / (3x - 2)²
= [15x - 10 - 15x - 12] / (3x - 2)²
= -22 / (3x - 2)²
2007-12-09 07:17:44 · answer #2 · answered by mountainpenguin 4 · 0⤊ 0⤋
5*(3x-2) - 3*(5x+4) \ (3x-2)^2
2007-12-09 07:14:52 · answer #3 · answered by Anonymous · 0⤊ 0⤋
5x+4=3x-2
5x-3x= -2-4
2x= -6
x = -6/2
x = -3
2007-12-09 07:12:18 · answer #4 · answered by Akhs 3 · 0⤊ 0⤋
f (x) = (5x + 4) / (3x - 2)
f `(x) is given by:-
(3x - 2) (5) - (5x + 4)(3)
-------------------------------
(3x - 2)²
15x - 10 - 15x - 12
------------------------
(3x - 2)²
- 22 / (3x - 2)²
2007-12-09 07:42:07 · answer #5 · answered by Como 7 · 2⤊ 0⤋
what if u use (v)(du/dx)-(u)(dv/dx) over v^2
and u will get -22/(3x-2)^2..
2007-12-09 07:16:26 · answer #6 · answered by Anonymous · 0⤊ 0⤋