Learning calculus

rockytriton · 04-18-2005

Hi everyone, I'm trying to teach myself calculus here and I'm having some algebra problems, I hope you can help me out here.

I'm having trouble seeing how I get from

[−1/(1 + h)^2] − [−1/1^2]
----------------------------
h

to

−1 − [−(1 + h)^2]
--------------------
h(1 + h)^2

I know that I should know this stuff before tackling calc but I remember things things all the time. Also I see that it goes from the above form to:

2h + h^2
------------------
h + 2h^2 + h^3

The bottom part I can understand as first doing (1 + h)^2 as 1 + 2h + h^2 and then multiplying each term by h, but I'm not sure on the top part what went on

Thanks in advance!

maddog · 04-18-2005

Quote:

Originally Posted by rockytriton

Hi everyone, I'm trying to teach myself calculus here and I'm having some algebra problems, I hope you can help me out here.

I'm having trouble seeing how I get from

[−1/(1 + h)^2] − [−1/1^2]
----------------------------
h

to

−1 − [−(1 + h)^2]
--------------------
h(1 + h)^2

I know that I should know this stuff before tackling calc but I remember things things all the time. Also I see that it goes from the above form to:

2h + h^2
------------------
h + 2h^2 + h^3

The bottom part I can understand as first doing (1 + h)^2 as 1 + 2h + h^2 and then multiplying each term by h, but I'm not sure on the top part what went on

Thanks in advance!

Rocky,

Your first formula

(-1/(1 + h)^2 - (-1/1^2)) / h first simplifies rewritting it a bit (by using distributive property)

-1/ h(1 + h)^2 + 1 / h Now make denominator look same by multiplying right term

by (1 + h)^2/(1+h)^2

(-1 + (1 + h)^2 ) / h(1 +h)^2 <-- to here by gathering terms in numerator

This gives the first interim solution multiplying out as

(-1 + 1 + 2h + h^2 ) / h(1 + 2h + h^2) =>

(2h + h^2) / (h + 2h^2 + h^3) <--- to get your answer.

However you can factor more to

(h^2 + 2) / (h + 1)^2 <-- to here.

Hope this helps.

Maddog

rockytriton · 04-19-2005

Ok, I think this is the part that I'm having problems with:

going from

-1/(1 + h)^2 - -1/1^2
----------------------------
h

to

-1
--------- + 1
h(1 + h)^2
---------------------
h

I understand how you get "- -1/1^2" to be + 1, but I don't understand how (1 + h)^2 changes to h(1 + h)^2.

Sorry, I guess I need a good refresher course on division with fractions and such... I really appreciate your help here.

maddog · 04-19-2005

Quote:

Originally Posted by rockytriton

[?1/(1 + h)^2] ? [?1/1^2]
----------------------------
h

to

?1 ? [?(1 + h)^2]
--------------------
h(1 + h)^2

Rocky,

Initially --> {[-1 / (1 + h)^2] - [ - 1/1^2] } / h <-- first I'll split in two for you

[ -1 / (1 + h)^2] / h - [ -1/1^2] / h <--- simplifying

-1 / h(1 + h)^2 + 1/h <-- now to get common terms I need to multiply second by 1

1 = (1 + h)^2 / (1 + h)^2 <-- so this comes out to

( -1 + (1 + h)^2 ) / h(1 + h)^2 = (-1 - ( - (1 + h)^2) ) / h(1 + h)^2

which is your second formula above.

I would recommend you go buy a Schaum Outline series on Algebra (NOT Abstract)
(you will not understand that). There are a lot of tricks in there you can use. ;D

maddog

TINNY · 04-20-2005

can anyone help me obtain y in terms of x here:

dy/dx = (y^2 + y)/(sin x)

I couldn't manage to use separable differentiation because i couldn't integrate 1/sin x

rockytriton · 04-20-2005

Quote:

Originally Posted by maddog

Rocky,

Initially --> {[-1 / (1 + h)^2] - [ - 1/1^2] } / h <-- first I'll split in two for you

[ -1 / (1 + h)^2] / h - [ -1/1^2] / h <--- simplifying

-1 / h(1 + h)^2 + 1/h <-- now to get common terms I need to multiply second by 1

1 = (1 + h)^2 / (1 + h)^2 <-- so this comes out to

( -1 + (1 + h)^2 ) / h(1 + h)^2 = (-1 - ( - (1 + h)^2) ) / h(1 + h)^2

which is your second formula above.

I would recommend you go buy a Schaum Outline series on Algebra (NOT Abstract)
(you will not understand that). There are a lot of tricks in there you can use. ;D

maddog

Thanks, I have a much better understanding now. I do have the Schuam Outline on College Algebra, maybe I should actually open it up. I bought it a few years ago while taking college algebra.

maddog · 04-20-2005

Quote:

Originally Posted by TINNY

can anyone help me obtain y in terms of x here:

dy/dx = (y^2 + y)/(sin x)

I couldn't manage to use separable differentiation because i couldn't integrate 1/sin x

To solve for (y^2 + y)/sin x ? Hmm, ok.

int(dy) = int((y^2 + y)/sin x * dx)

You will need a substitution where y = f(x) or x = g(y) to proceed... hmmm

int((1/(y^2 + y)dy) = int((1/sin x)dx) <-- no I need no substitution as before. Instead I need to
simply by using substitution.

The right side can be found in a book on calculus (I'm a bit rusty). The left you can do by simple
substitution so try it out.

maddog

Bo · 04-21-2005

in my integral books there is no exact olution for 1/sin(x) (only a recursive solution)...
I also cant find a better solution, so what do you need this for tinny? Things would get much simpeler if one can take sin(x)~x for example

Bo