P(t) = 50(2)^(t/2)
$\displaystyle {\mathrm{d}\over\mathrm{d}t}(50\cdot 2^{t/2}) = 50 {\mathrm{d}\over\mathrm{d}t} (2^{t/2}) = 50{\mathrm{d}\over\mathrm{d}t} (e^{\log 2})^{t/2} = $ $\displaystyle 50{\mathrm{d}\over\mathrm{d}t} e^{\frac12t\log 2} = 50 e^{\frac12t\log 2} {\mathrm{d}\over\mathrm{d}t}(\frac12 t \log 2) = 50\cdot 2^{t/2} \cdot \frac{\log 2}2 = 25\cdot 2^{t/2}\log 2$
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?