Answer
(a) $c^{'}(t)= (-\frac{kA}{V}) (c_0-C) e^{-\frac{kAt}{V}} $
(b) See explanation
Work Step by Step
$c^{'}(t)=\frac{kA}{V}[C-c(t)]$ ........................ (1)
$c(t)=(c_0-C)e^{-\frac{kAt}{V}}+C$ .............................. (2)
(a)
Taking derivative of equation (2)
$c^{'}(t)=\frac{d( (c_0-C)e^{-\frac{kAt}{V}}+C )}{dt}$
$c^{'}(t)=\frac{d( (c_0-C)e^{-\frac{kAt}{V}}}{dt}+\frac{dC}{dt}$
$c^{'}(t)=(c_0-C)\frac{de^{-\frac{kAt}{V}}}{dt}+0$
$c^{'}(t)=(c_0-C) e^{-\frac{kAt}{V}} \frac{d ( -\frac{kAt}{V} )}{dt}$
$c^{'}(t)=(c_0-C) e^{-\frac{kAt}{V}} (-\frac{kA}{V}) \frac{d ( t )}{dt}$
$c^{'}(t)= (-\frac{kA}{V}) (c_0-C) e^{-\frac{kAt}{V}} $
(b)
Putting $c^{'}(t)= (-\frac{kA}{V}) (c_0-C) e^{-\frac{kAt}{V}} $ in equation (1)
$ (-\frac{kA}{V}) (c_0-C) e^{-\frac{kAt}{V}} =\frac{kA}{V}[C-c(t)]$
Putting $t=0, c(0)=c_0$
$ (-\frac{kA}{V}) (c_0-C) e^{-\frac{kA(0)}{V}} =\frac{kA}{V}[C-c(0)]$
$ (-\frac{kA}{V}) (c_0-C) =\frac{kA}{V}[C-c_0)$
$ - (c_0-C) =C-c_0)$
$ C-c_0 =C-c_0$
LHS=RHS
Hence equation (2) is correct antiderivative of equation (1)
