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The chain rule strikes again!
thats the bit i dont get though.
Whoops, I wrote the solution, not the problem. What I meant was:
Sure, Mathsy it would. If p had a coefficient of a then the answer would be -a/(1-ap)²
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:
Surely the coefficient of p doesn't matter?
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).
Ah, sorry, should have probably been clearer.
I think Ricky's rhyme is referring to the quotient formula that irspow posted earlier.
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?
Low d' Hi minus Hi d' Low, all over the denominator squared we go.
thanks thats very helpful.
They are asking you to differentiate 1/(1-p) with respect to p.
how does this work? thanks