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Ricky
2006-01-16 02:45:36

The chain rule strikes again!

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

eddie
2006-01-16 02:25:59

#### 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
2006-01-15 11:45:51

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

(1-p)-¹

irspow
2006-01-15 11:10:29

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

Ricky
2006-01-15 10:51:57

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.

mathsyperson
2006-01-15 10:27:20

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

irspow
2006-01-15 10:17:01

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).

Ricky
2006-01-15 05:07:11

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.

mathsyperson
2006-01-15 04:56:43

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.

eddie
2006-01-15 04:43:36

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!

Ricky
2006-01-15 04:22:25

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

eddie
2006-01-15 04:03:40

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

thank you

irspow
2006-01-15 03:35:10

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

eddie
2006-01-15 03:20:39

how does this work? thanks

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