Question

f(t)=At^b e^(-ct) Show that the rate of change in concentration with respect to time is f'(t)=A(b-ct)t^b-1 e^(-ct)

Answer 1

# Question # 11279

$f(t) = At^{b}e^{-ct}$ Show that the rate of change in concentration with respect to time is $f'(t) = A(b - ct)t^{b - 1}e^{-ct}$.

Solution. By definition, the rate of change of $f$ is $f'$, using the formula of differentiating of product, one can get that $f'(t) = Abt^{b-1}e^{-ct} - cAt^b e^{-ct} = A(b - ct)t^{b-1}e^{-ct}$.

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