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## #1 2006-01-15 03:20:39

eddie
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### how to differente this?

how does this work? thanks

d/dp 1/(1-p) = 1/(1-p)^2

## #2 2006-01-15 03:35:10

irspow
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### Re: how to differente this?

They are asking you to differentiate 1/(1-p) with respect to p.

With fractions differentiation works like this;

1. Take the derivative of the numerator and multiply it by the divisor.
2. Take the derivative of the divisor and multiply it by the negative of the numerator.
3. Add 1. and 2. together and divide this by the divisor squared.

or... the derivative of f(x)/g(x) = [f'(x)g(x) - f(x)g'(x)] / g(x)²

In you above equation;  (by the steps above)

[0(1-p) - 1(-1)] / (1-p)² = 1/(1-p)²

Last edited by irspow (2006-01-15 03:36:06)

## #3 2006-01-15 04:03:40

eddie
Guest

### Re: how to differente this?

one more quick question, which is the correct way to write the second derivative? d2/d2p or d2/dp2

thank you

## #4 2006-01-15 04:22:25

Ricky
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### Re: how to differente this?

Low d' Hi minus Hi d' Low, all over the denominator squared we go.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #5 2006-01-15 04:43:36

eddie
Guest

### Re: how to differente this?

lol nice rhyme but i have no idea what you're talking about! i thought i just needed to know if the ^2 in the denominator goes after the d or after the variable. whats hi d and low d?

ps. i like the quote in your signature!

## #6 2006-01-15 04:56:43

mathsyperson
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### Re: how to differente this?

I think Ricky's rhyme is referring to the quotient formula that irspow posted earlier.

In answer to your question, the ^2 goes after the variable on the denominator.

Why did the vector cross the road?
It wanted to be normal.

## #7 2006-01-15 05:07:11

Ricky
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### Re: how to differente this?

Ah, sorry, should have probably been clearer.

For the first question, it's much easier to do:

1/(1-p)^2 = (1-p)^-2

But either way works.  And besides, it's much better know multiple ways to differentiate.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #8 2006-01-15 10:17:01

irspow
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### Re: how to differente this?

Ricky, that is correct but only because the coefficient of p is a negative one.  I was just pointing out the actual steps for any type of integration of the form f(x)/g(x).

## #9 2006-01-15 10:27:20

mathsyperson
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### Re: how to differente this?

Surely the coefficient of p doesn't matter?
Isn't it because the numerator in this case is a constant?

Why did the vector cross the road?
It wanted to be normal.

## #10 2006-01-15 10:51:57

Ricky
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### Re: how to differente this?

You do have f(x)/g(x), but you also have k/g(x).  Like I said, either way works.  You could also multiply them:

f(x) * g-¹(x), and use the product rule.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #11 2006-01-15 11:10:29

irspow
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### Re: how to differente this?

Sure, Mathsy it would.  If p had a coefficient of a then the answer would be -a/(1-ap)²

## #12 2006-01-15 11:45:51

Ricky
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### Re: how to differente this?

Whoops, I wrote the solution, not the problem.  What I meant was:

(1-p)-¹

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #13 2006-01-16 02:25:59

eddie
Guest

### Re: how to differente this?

#### Ricky wrote:

Whoops, I wrote the solution, not the problem.  What I meant was:

(1-p)-¹

thats the bit i dont get though.

why isnt the differential -1(1-p)^-2, so -1/(1-p)^2

## #14 2006-01-16 02:45:36

Ricky
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### Re: how to differente this?

The chain rule strikes again!

Using the chain rule, the derivative of 1 - p is -1.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."