What Is the Derivative of the Function $ s \left( t \right) = \left( 1 - {e}^{\frac{-t}{RC}} \right) u \left( t \right) $?
By the Product Rule (With abuse of the derivative of Unit Step):

$$\begin{align*}
\frac{d}{dt} \left[ \left( 1 - {e}^{\frac{-t}{R C}} \right) u \left( t \right) \right] &amp; = \frac{d}{dt} \left[ \left( 1 - {e}^{\frac{-t}{R C}} \right) \right] u \left( t \right) + \left( 1 - {e}^{\frac{-t}{R C}} \right) \frac{d}{dt} u \left( t \right) \\
&amp; = \frac{1}{R C} {e}^{\frac{-t}{RC}} u \left( t \right) + \left( 1 - {e}^{\frac{-t}{R C}} \right) \delta \left( t \right)
\end{align*}$$

By the properties of Dirac's Delta $ f \left( t \right) \delta \left( t \right) = f \left( 0 \right) \delta \left( t \right) $ we get:

$$ \frac{1}{R C} {e}^{\frac{-t}{RC}} u \left( t \right) + \left( 1 - {e}^{\frac{-t}{R C}} \right) \delta \left( t \right) = \frac{1}{R C} {e}^{\frac{-t}{RC}} u \left( t \right) + 0 \delta \left( t \right) = \frac{1}{R C} {e}^{\frac{-t}{RC}} u \left( t \right) $$

Remark
The above is with some abuse of Math (Derivative of Unit Step as Delta Function, Zero multiplied by Delta is zero, etc...) which is used (Again, it is abuse but usually gets to the right place).
If I remember correctly, even this intuitive abuse is used in Desor.