hey, this is a problem from my econ hw.
I need to derive q= 20/(1+p) with respect to p. im just not quite sure how to handle the (1+p) part.
I do know that when i derive q with respect to p in this equation, q= 20/p , that you can make the equation into q= 20p^-1 making it easy to do and getting an answer of -20(p^-2) .
so can i do the first equation like the second one by taking q= 20(1+p)^-1 ? and the answer would be -20(1+p)^-2 . or does the 1 go away after you derive it, so the answer would just be 20/(p^2) ? or am i completely off?
any help or incite would be appreciated
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my derivative skills are shaky but i think you're on the right track.
if you let (1+p) = r, then you have
q= 20r^-1
dq= -20r^-1dr
and since r = p+1
dr= 1
so you'd have -20(1+p)^-2 a.k.a. -20/((1+p)^2)
i think.xDOLPHINSx likes this. -
i'll go with that answer. thanks guys.
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No problem. I remember hating calculus in Econ classes. If you can use one, I'd highly recommend a TI-89. Does all the derivatives and integrals for you. Bit of a battery eater but it's worth it in my opinion. Saved my *** when I could use it.
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I could never use a graphical calculator in my 1st two calc classes at Toledo, then at Akron, I went through calc 2 through differential equations without a graphing calculator. I was the only person in chemical engineering in my grad class that didn't use one (I don't even know how, lol).