Question

who can help me to determine the derivative by limit..please

Answer 1

\[f(x)=2x-1 \over 3x-5\]

Answer 2

by limit i believe is also called first principles ..

Answer 3

...and this problem is more for helping you practice the long way of finding a derivative so that you appreciate the shortcuts later on

Answer 4

help me please..

Answer 5

what have you got so far?

Answer 6

I make that like this \[[(2(x+\Delta x)-1 \over 3(x+\Delta x)+] - 2x-1 \over 3x+5\]

Answer 7

\[f(x)=\frac{2x-1}{3x-5}\]
\[\lim_{h->0} \frac{\cfrac{2(x+h)-1}{3(x+h)-5}-\cfrac{2x-1}{3x-5}}{h}\]

Answer 8

the idea is to expand it out; get like denominators on top and algebra it into something we can use ....

Answer 9

\[\lim_{h->0} \frac{\cfrac{2(x+h)-1}{3(x+h)-5}-\cfrac{2x-1}{3x-5}}{h}\]
\[\lim_{h->0} \frac{\cfrac{2x+2h-1}{3x+3h-5}*\cfrac{(3x-5)}{(3x-5)}-\cfrac{2x-1}{3x-5}*\cfrac{(3x+3h-5)}{(3x+3h-5)}}{h}\]
\[\lim_{h->0} \frac{\cfrac{(6x^2 +6xh -13x -10h+5) -(6x^2 +6xh -13x -3h+5)}{9x^2 +9xh -30x -15h +25}}{h}\]
\[\lim_{h->0} \frac{\cfrac{6x^2 +6xh -13x -10h+5 -6x^2 -6xh +13x +3h-5}{9x^2 +9xh -30x -15h +25}}{h}\]
\[\lim_{h->0} \frac{\cfrac{ -7h}{9x^2 +9xh -30x -15h +25}}{h}\]
\[\lim_{h->0} \frac{\cfrac{ -7h}{9x^2 +9xh -30x -15h +25}}{h}*\cfrac{\cfrac{1}{h}}{\cfrac{1}{h}}\]
\[\lim_{h->0} \cfrac{-7}{9x^2 +9xh -30x -15h +25}\]

now when h=0 we get

\[\cfrac{-7}{9x^2 +9x(0) -30x -15(0) +25}\]
\[\cfrac{-7}{9x^2 -30x +25}\]
if i did it right

Answer 10

(2x−1)/(3x−5) the short way is to use the quotient rule:
[t/b]' = (bt'-b't)/b^2

(3x-5)(2) - 3(2x-1)
----------------
(3x-5)^2


6x-10 -6x +3 -7
--------------- = -------------
9x^2 -30x +25 9x^2 -30x +25

Answer 11

but for our assignment we need to use the three way step..

Answer 12

i got no idea what a "3way step" is :)

Answer 13

the f(x+ Deltax)-f(X) after you got the answer next is you will going to divide it into delta x and the last is the limit...

Answer 14

all I did was rename "delta x" as "h"

Answer 15

ah..okay...

Answer 16

i worked the f(x+h)-f(x) and then limited it ... so that covers it :)

Answer 17

can you try try this one...\[-x ^{3}-x ^{2}\]

Answer 18

id rather you try it try it an then I can see what you know

Answer 19

oh...amistre64 its positive 5 not negative..

Answer 20

you mean 3x+5?

Answer 21

yes..

Answer 22

doesnt matter; the process is the same regardless; just the numbers change a little

Answer 23

okay...thank you by the way..

Answer 24

are you using cross multiplication in the second step?

Answer 25

the second step I determined "like denominators" so that I could actually subtract f(x) from f(x+h)

Answer 26

okay..

Answer 27

cross multiply would be a bad description, if anything I multiplied the denominators together :) ... bottom multiplied perhaps?

Answer 28

yes...i also think that:)

Answer 29

amistre in these equation: \[-x ^{3}-x ^{2}\]

Answer 30

do i need to write in this form -(x-deltax) or (-x-deltax)