What is the derivative of the integral of a function? Is this correct ?
$$
\frac{d}{dt} \left( \int_0^t \phi(t)dt \right) = \phi(t)
$$
If not, how can I recover $$ \phi(t) $$ knowing only $$ \int_0^t \phi(t)dt $$ ?
 A: The expression $\int_{0}^{t}\phi\left(t\right){\rm d}t$ makes no
sense to me. It cannot be that $t$ is a variable and a constant at
the same time. I presume that you mean $\int_{0}^{t}\phi\left(x\right){\rm d}x$ 
For $t>0$ prescribe $\phi(t)$ by $t\mapsto 0$ if $t\neq 1$ and $t\mapsto c$ otherwise, where $c$ is some constant. 
Then $\Phi\left(t\right)=\int_{0}^{t}\phi\left(x\right){\rm d}x=0$ for
each $t$ so that $\Phi'\left(t\right)=0$ for each $t$. 
However,
if $c\neq0$ then $\Phi'\left(1\right)=0\neq c=\phi\left(1\right)$. 
This can be done with every $c$ so apparantly $\phi(1)$ cannot be recovered.
Things get better if you demand $\phi(t)$ to be continuous.
A: (i) No, it is not correct. You probably mean 
$$
\frac{\text{d}}{\text{d}t} \left( \int_0^t \phi(x) \, \text{d}x \right) = \phi(t)
.$$
This  equality is also known as the Fundamental Theorem of Calculus, Part 1.
(ii) You can recover $\phi (t)$ by differentiating, as demonstrated above.
A: Let $\Phi$ the antiderivative of $\phi$ ($\Phi'(t)=\phi(t)$) such as $\Phi(0)=0$ 
We have $\Phi(t)=\int_0^t \phi(t)dt$ thus $\dfrac{d}{dt}\left(\int_0^t \phi(t)dt\right)=\dfrac{d\Phi}{dt}=\phi(t)$
