#1

Right, so I have some Calculus HW and Im trying to figure out how to get the common denominator in order to get the derivative of the original equation. Problem is I forgot how to get the common denominator.

Original function: f(x)= (x^2+1)/(x-2). So after using the Lagrange method i got to:

[( (a+h)^2+1) / (a+h)-2] - [(a^2+1)/(a-2)] both of those brackets over h. SO you got a divition on top of another divition, and theres were im confused. So could anyone help me get the common denominator or show me how to do it?

Thanks

Rulfo

Original function: f(x)= (x^2+1)/(x-2). So after using the Lagrange method i got to:

[( (a+h)^2+1) / (a+h)-2] - [(a^2+1)/(a-2)] both of those brackets over h. SO you got a divition on top of another divition, and theres were im confused. So could anyone help me get the common denominator or show me how to do it?

Thanks

Rulfo

#2

Right, so I have some Calculus HW and Im trying to figure out how to get the common denominator in order to get the derivative of the original equation. Problem is I forgot how to get the common denominator.

Original function: f(x)= (x^2+1)/(x-2)divided by 0.So after using the Lagrange method i got to:

[( (a+h)^2+1) / (a+h)-2] - [(a^2+1)/(a-2)] both of those brackets over h. SO you got a divition on top of another divition, and theres were im confused. So could anyone help me get the common denominator or show me how to do it?

Thanks

Rulfo

#3

x/y/z = z.x/y

#4

I hate math and I haven't done it in 5 years. I was actually going to shout at you for not doing your own homework but after reading your thread, you've actually set a prime example of how to ask for homework help, if ever, in the PIT. Well done for making the first homework thread without asking for the answers directly and also having done a bit of work before hand.

However, UG still isn't the correct place for homework advice. I suggest you log onto The Student Forum if you really can't get to a teacher before it's handed in or if you don't have any friends to call for help.

However, UG still isn't the correct place for homework advice. I suggest you log onto The Student Forum if you really can't get to a teacher before it's handed in or if you don't have any friends to call for help.

*Last edited by Harmonius at Nov 26, 2009,*

#5

WTF is the Lagrange method?

Can't you just get the derivative from the original function using the quotient rule?

Can't you just get the derivative from the original function using the quotient rule?

#6

Can't you just get the derivative from the original function using the quotient rule?

I agree

#7

The problem isn't the actual work of getting the derivative, the problem is I forgot how to get a common denominator, so i can't finish the problem until i finish the common denominator. So if anyone could point me to the right direction on how to get the common denominator for a thing like that i would really, really appreciate it.

I hate recurring to the Pit for stuff like this, but I lost my math book and I really need this done for tomorrow. And no, its not dividing by zero...

I hate recurring to the Pit for stuff like this, but I lost my math book and I really need this done for tomorrow. And no, its not dividing by zero...

#8

WTF is the Lagrange method?

Can't you just get the derivative from the original function using the quotient rule?

I mistyped that, it's not lagrange method, is the lagrange notation f ' (x) instead of the D(x)/d(x). And what is the quotient rule. My math teacher kinda fails at teaching so I dont know what the quotient rule is, so if anyone could show me how to do it

Thanks again.

#9

Multiply by (a+h)-2 and (a-2) so that it becomes:

[(a+h)^2+1](a-2) - (a^2+1)[(a+h)-2] over h[((a+h)-2)(a-2)]

Multiply out and collect like terms and cancel out h's

EDIT: Scratch that.

If you're gonna use the quotient method then let

f=(x^2+1) and g=(x-2).

Then do f'g-g'f divdided by g^2

[(a+h)^2+1](a-2) - (a^2+1)[(a+h)-2] over h[((a+h)-2)(a-2)]

Multiply out and collect like terms and cancel out h's

EDIT: Scratch that.

If you're gonna use the quotient method then let

f=(x^2+1) and g=(x-2).

Then do f'g-g'f divdided by g^2

*Last edited by dma529 at Nov 26, 2009,*

#10

I'm horrible at math

And I mean horrible.

But can't you just get the common denominator by multiplying both denominators by eachother?

And I mean horrible.

But can't you just get the common denominator by multiplying both denominators by eachother?

#11

And that gives me the actual derivative (f prime of x)?

Ill try that out.

Thanks dma

Ill try that out.

Thanks dma

#12

I'm horrible at math

And I mean horrible.

But can't you just get the common denominator by multiplying both denominators by eachother?

Nop, because its a division on top of another division. And its not like its two fractions. Theres a thing called the flip law or something like that that helps you get the common denominator but I forgot.

#13

I'd just use the quotient rule if I were you because it's a lot less work than the original way you had it.

#14

In future please use the math/science thread.

#15

Multiply by (a+h)-2 and (a-2) so that it becomes:

[(a+h)^2+1](a-2) - (a^2+1)[(a+h)-2] over h[((a+h)-2)(a-2)]

Multiply out and collect like terms and cancel out h's

EDIT: Scratch that.

If you're gonna use the quotient method then let

f=(x^2+1) and g=(x-2).

Then do f'g-g'f divdided by g^2

Cool, that seems easier than the first one

Thanks

To King

Yea i will use the thread, i just really needed the info

#16

a/x + b/y = ay/xy + bx/xy