Using the chain rule, we have
P ′ ( t ) = C ( e k t ) ′ = C k e k t {\displaystyle P'(t)=C(e^{kt})'=Cke^{kt}} and hence P ′ ( 0 ) = C k {\displaystyle P'(0)=Ck} .
Recall that we find C = 300 e 3 5 ln 6 {\displaystyle C={\frac {300}{e^{{\frac {3}{5}}\ln 6}}}} and k = ln 6 25 {\displaystyle k={\frac {\ln 6}{25}}} in part (a).
Therefore, P ′ ( 0 ) = 300 e 3 5 ln 6 ⋅ ln 6 25 = 12 ln 6 e 3 5 ln 6 {\displaystyle \color {blue}P'(0)={\frac {300}{e^{{\frac {3}{5}}\ln 6}}}\cdot {\frac {\ln 6}{25}}={\frac {12\ln 6}{e^{{\frac {3}{5}}\ln 6}}}} .
and hence 
.
and 
in part (a).
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