2007-03-22 06:52:08 · 8 answers · asked by nala 1 in Science & Mathematics ➔ Engineering
(2)(4)(2x-3)^3/(3)(-5)(3x-2)^6
-8(2x-3)^3/15(3x-2)^6
Maybe. I am 10 years removed from calculus.
2007-03-22 06:57:24 · answer #1 · answered by Jimee77 4 · 0⤊ 0⤋
D((2x-3)^4/(3x-2)^5)= D((2x-3)^4 * (3x-2)^-5) // **** the quotient rule use the product one.
(2x-3)^4*D[(3x-2)^-5] + (3x-2)^-5*D[(2x-3)^4]
note; D[({3x-2})^-5]= -5({3x-2})^-6*{// D({3x-2})] //=-5({3x-2})^6 //3dx// *which is the derivititve of {3x-2}.
Now just do that with the other one, and combine like terms. I like to think of a derivitives like this.
D[ u^n]= (n) *{u}^(n-1) d({u}). Just think of the Dy/dx =something with the dx moved over to the other side. Now whatever you u is then you can just take the derivitive untill you get what you are taking it with respect to like if u={3x-2} and you wanted your derivitive in terms of dx then you just have to go one step further... d{3x-2}=3*dx.
2007-03-22 07:25:22 · answer #2 · answered by Michael M 4 · 0⤊ 0⤋
Use quotient rule
let f(x) = u/v
f `(x) = [(v).(du/dx) - (u).(dv/dx)] / v²
u = (2x - 3)^(4)
du/dx = 4(2x - 3)^(3).2 = 8.(2x - 3)^(3)
v = (3x - 2)^(5)
dv/dx = 5.(3x - 2)^(4).3 = 15.(3x - 2)^(4)
v² = (3x - 2)^(10)
Consider numerator (N) (to assist typing !):-
N is given by:-
8.(3x - 2)^(5).(2x - 3)³ - 15.(2x - 3)^(4).(3x - 2)^(4)
1 step of work in here (no room!) leads to:-
N = (2x - 3)³.(3x - 2)^(4)[29 - 6x]
D = (3x - 2)^(10)
I = N/D
I = (2x - 3)³ (29 - 6x) / (3x - 2)^(6)
Comment:Typing a nightmare but method is OK----- please check.
2007-03-22 08:56:47 · answer #3 · answered by Como 7 · 0⤊ 0⤋
Use the quotient rule, combined with the chain rule. (bottom)(derv of top) - (top)(derv of bottom)/ (square the bottom)
(3x-2)^5(4(2x-3)^3(3))-(2x-3)^4(5(3x-2)^4(3)) / (3x-2)^7
2007-03-22 07:22:51 · answer #4 · answered by shavger_r 2 · 0⤊ 0⤋
use the product rule for this (x)[?(2x+a million)] so first take the spinoff of the 1st term expanded by using the 2nd term: a million[?(2x+a million)] plus the 1st term expanded by using spinoff of the 2nd term: (x)(a million/2)[(2x+a million)^(-a million/2)](2) in case you simplify then you certainly might desire to come out with ?(2x+a million)+x/[?(2x+a million)]
2016-11-27 22:38:31 · answer #5 · answered by ? 4 · 0⤊ 0⤋
product rule
(2x-3)^4*(3x-2)^-5
u^n*v^i
(du*n)u^(n-1)*v^i + (dv*i)v^(i-1)*(u^n)
8*(2x-3)*(3x-2)^-5 -15(3x-2)^-6*(2x-3)^4
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2007-03-22 06:57:56 · answer #6 · answered by Anonymous · 0⤊ 0⤋
stop trying to get people to do your homework.
2007-03-22 06:54:13 · answer #7 · answered by Michelle118 4 · 0⤊ 0⤋
x=Wal-Mart.
2007-03-22 06:55:07 · answer #8 · answered by double D 2 · 0⤊ 0⤋