Following are the piece wise polynomial function for input plasma
$$ C_a(t) = \begin{cases}0& t\leq t_d\\ \displaystyle\sum_{n=1}^3\frac{a_n}{t_\max-t_d}(t_-t_d)& t_d\leq t\leq t_\max\\ \displaystyle\sum_{n=1}^3a_ne^{-b_n(t-t_\max)}& t_\max\leq t\end{cases}$$
where C_a is input plasma function
$$C_{tot}(t) = (\frac{k_1k_2}{k_2+k_3}\cdot e^{-(k2+k3)t}+\frac{k_1k_3}{k_2+k_3})\bigotimes C_a(t).$$
C_tot is total Concentration output
I would like to perform Differentiation of convolution integral $$C_{tot}(t)$$with respect to $$k3(t)$$
Please Let me know how i can proceed with the following computation of differentiation of Convolution integral
I think your problem is simpler than you're making it. Let's start with the basic convolution integral: $$h(t) = \int f(t)g(t-\alpha)d\alpha $$
From our discussion it is evident that $k_3$ is constant wrt $t$. So I will answer based on this assumption. If this is true than you need only differentiate the unnamed part of your equation, let's call this $f(t)$. Since there is no $k_3$ in $C_a(t)$ it can be omitted. To illustrate:
$$\frac{dC_{tot}}{dk_3} = \frac{d}{dk_3} \int f(t,k_3)g(t-\alpha)d\alpha$$ is equivalent through basic properties to $$\frac{dC_{tot}}{dk_3} = \int \frac{d}{dk_3} [f(t,k_3)]g(t-\alpha)d\alpha$$
So all we need to find is $\frac{d}{dk_3} [f(t,k_3)]$ which is $$\frac{d}{dk_3} [(\frac{k_1k_2}{k_2+k_3}⋅e^{−(k_2+k_3)t}+\frac{k_1k_3}{k_2+k_3})]$$ Break it into two terms: $$\frac{d}{dk_3}[\frac{k_1k_2}{k_2+k_3}⋅e^{−(k_2+k_3)t}] + \frac{d}{dk_3}[\frac{k_1k_3}{k_2+k_3}]$$
Product rule on the first term
$$\frac{d}{dk_3}[\frac{k_1k_2}{k_2+k_3}⋅e^{−(k_2+k_3)t}] = \frac{k_1k_2}{k_2+k_3}(-t)e^{-(k_2+k_3)t}+e^{-(k_2+k_3)t}\frac{-k_1k_2}{(k_2+k_3)^2}$$ $$\frac{d}{dk_3}[\frac{k_1k_2}{k_2+k_3}⋅e^{−(k_2+k_3)t}] = \frac{-k_1k_2(k_2t+k_3t+1)}{(k_2+k_3)^2}e^{-(k_2+k_3)t}$$
Quotient rule on the second term: $$\frac{d}{dk_3}[\frac{k_1k_3}{k_2+k_3}] = \frac{k_1k_2}{(k_2+k_3)^2}$$
Put it all back together: $$\frac{df}{dk_3} = \frac{-k_1k_2(k_2t+k_3t+1)}{(k_2+k_3)^2}e^{-(k_2+k_3)t} + \frac{k_1k_2}{(k_2+k_3)^2}$$
And recall that $$\frac{dC_{tot}}{dk_3} = \frac{df}{dk_3} \bigotimes C_a(t)$$
If you wanted to minimize $f$ you would set $\frac{df}{dk_3}=0$ and solve. You would get: $$(k_2t+k_3t+1)e^{-(k_2+k_3)t} = 0$$ $$k_3=k_2-\frac{1}{t}$$