I have an equation that I want to differentiate with respect to time. The integral is what confuses me.
The equation is: $$\int_0^t e^{-C(t-\tau)}f(\tau) \, d\tau$$
I have the answer and I think it is correct but I am not sure about the logic applied to get to the answer. The answer is: $$e^{-C(t-t)}f(\tau) - \int_0^t Ce^{-C(t-\tau)}f(\tau) \, d\tau$$
The first term is just $f(\tau)$.
Please let me know about the method used here and also point out any error if I had made anywhere.
$$\int_{0}^{t}e^{-C(t-\tau)}f(\tau)d\tau$$ is a function of $t$. By the generalized fundamental theorem of calculus (Liebniz's rule) we get $$ f(t)+\int_0^t\frac{d}{dt}e^{-C(t-\tau)}f(\tau)\mathrm d\tau=f(t)-C\int_0^te^{-C(t-\tau)}f(\tau)\mathrm d\tau $$ where the first term is evaluating the integrand at $t=\tau$ and the second is the correction term since the integrand is also a function of $t$.
This is a particular case of Leibniz integral rule for differentiating an integral.
$${\frac {\mathrm {d} }{\mathrm {d} t}}\left(\int _{a(t)}^{b(t)}f(x,t)\,\mathrm {d} x\right)=\int _{a(t)}^{b(t)}{\frac {\partial f}{\partial t}}\,\mathrm {d} x\,+\,f{\big (}b(t),t{\big )}\cdot b'(t)\,-\,f{\big (}a(t),t{\big )}\cdot a'(t)$$
One might recognize this to be a combination of the multivariate chain rule and the fundamental theorem of calculus.