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
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.
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.
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).
