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**eddie****Guest**

how does this work? thanks

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

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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 (2006-01-14 04:36:06)*

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**eddie****Guest**

thanks thats very helpful.

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

thank you

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

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**eddie****Guest**

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!

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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).

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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

(1-p)-¹

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**eddie****Guest**

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

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

The chain rule strikes again!

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

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