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

]]>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

]]>(1-p)-¹

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

]]>Isn't it because the numerator in this case is a constant?]]>

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

]]>ps. i like the quote in your signature!

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

thank you

]]>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

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

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