I want to work on a quick math program but i can't remember the formula for getting the derivative of a fraction such as

Lim X --> 0 X^4/ X-1

Any help from the calculas kings appreciated.

:alright:
Posted on 2001-12-28 14:15:20 by titan
As I remember it, for any fraction:

``````
F(x)  =  N(x)  / D(x)

May be evaluated by:

d N(x)|        d D(x)|
F(Xo) = -------|   /   -------|
dx    |        dx    |
|Xo            |Xo
``````

IE, the derivative of the numerator, divided by the derivative of the denominator, evaluated at the point of interest.
Posted on 2001-12-28 15:04:06 by Ernie
To check your results:
Posted on 2001-12-28 18:20:10 by eet_1024
Hi !

In general:

If you have two functions, g(x) and h(x), and g'(x) is the deriviation of g by x and h'(x) of h(x) by x then the basic rules are:

f(x) = g(x) + h(x) => f'(x) = g'(x) + h'(x)

f(x) = g(x) * h(x) => f'(x) = g'(x) * h(x) + g(x) * h'(x)

f(x) = g(x) / h(x) => f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x)^2)

f(x) = g(h(x)) => f'(x) = (dg / dh) * h'(x)

Greetings, CALEB
Posted on 2001-12-28 19:02:12 by Caleb
hi all,
Limit is not like derivative, The derivative is a special case of the limit.
The derivative is a limit like this:
lim h->0 (F(X+H)-F(X))/h
But limits in general differs.
There are some laws for it:
1- by using direct substitution like in your case:
lim x->0 X^4/X-1=0^4/0-1=0/-1=0
Therefor the limit=0
2-If the limit gives 0/0 by direct substitutions there will be other ways to deal with the limit
Posted on 2001-12-29 01:12:11 by amr
eet_1024: you do you create this equations? is there a program i can create such equations easily?

nop
Posted on 2001-12-29 08:17:12 by NOP-erator
it equals to 0 becuase when x -> 0 . image x = 0.0000000001
and when you will do x^4 the number will get lower and lower .
in your equation the numerator equals to zero.

but if your x-> 1 . you were in a problem becuase the denominator were equals to zero( and this is another story)

check out

nice math program

math editor

p.s
i'm sorry for my english and hope you understood my explantion
Posted on 2001-12-29 09:53:26 by eko
Maybe I should have given the complete answer.

For limit problems such as this, it is a legitimate technique to see 'how fast' the numerator and denominator get to the point. This is why the lim x->0 sin(x)/x =1 (cos(0)/1 = 1/1 = 1)

Anyway,

d
--- x^4 = 4x^3
dx

d
-- x-1 = 1
dx

thus 4x^3/1| = 0/1 = 0
|
|x=0

For the limit as x->1, this would evaluate to 4.
Posted on 2001-12-29 10:33:57 by Ernie
Hi Ernie !

Your way is the rule by l'Hospital but you can use it only if the fraction of the two functions is undefined (either 0/0 or oo / oo )
So you can use it for

lim x->1 (x^4/(x-1))

but not to get

lim x->0 (x^4/(x-1))

The rule itself says if you have lim x->x0 ( f(x) / g(x) ) and its form is ( 0/0 ) or ( 00 / 00 ) then

lim x->x0 ( f(x) / g(x) ) = lim x->x0 ( f'(x) / g'(x) )

if it exists !

Greetings, CALEB
Posted on 2001-12-29 11:20:36 by Caleb
You can use l'Hopital wherever you want. It helps to get you out of situations where the limit is undefined, but its not limited to those situations.

But in this case theres no need, don't forget the rule of thumb; if the limit makes sense, use it.
So Lim X --> 0 X^4/ X-1 becomes 0^4 / 0-1 = 0 / -1 = 0
Posted on 2001-12-29 12:17:46 by E¾in
Hi E?in !

You get wrong ! You cannot use l'Hospital everywhere you want to get the right solution ! It is only for those undefined limits. But you can transform terms like oo - oo or 0^0 to a fraction of functions and then use l'Hospital !

Greetings, CALEB
Posted on 2001-12-29 13:29:17 by Caleb
nop:
Plain and cheap paint.exe :)
Posted on 2001-12-29 17:20:22 by eet_1024