P(t) = 50(2)^(t/2)
Hmm, part of what i'm doing asks me to find the slope at 7 days; the answer is 196. However when I sub in 7 for t this is what I get:
P(t) = 50(2)^(t/2)
P'(t) = 25(2^(t/2)log2)
P'(7) = 25(2^(7/2)log2)
= 25((11.3137085)(0.3010299957)) *2^(7/2)=11.3137085, log2=0.3010299957
=25(3.40576562)
=85.14414051
The answer is 196, so i'm assuming I did something wrong. Perhaps is your derivative wrong, or did I do something wrong with my calculations?