I am trying to find the derivative of the following function:
$$f(t) = s - (m\cdot g/k)\cdot t+\left(m^2\cdot g/k^2\right)\cdot (1-\exp{(-k\cdot t/m)})$$
where
$$g = 9,81\\ m = 0.1\\ k = 0.1488\\ \text{and }s = 68.$$
I assumed that because of knowning so many variables, I theoretically have a lot of constants, and thus came to the derivative:
$$f'(t) = 1 + (k/m) \cdot a \cdot \exp{(-k\cdot t/m)}$$
Is this the correct derivative?
\begin{align}f(t) &= s - \left(\frac{mg}k\right)t+\frac{m^2g}{k^2}\left(1-\exp\left(-\frac{kt}m\right)\right) \\ &=s - \left(\frac{mg}k\right)t+\frac{m^2g}{k^2}-\frac{m^2g}{k^2}\exp\left(-\frac{kt}m\right) \\\end{align}
The first and third term is independent of $t$, the second term is linear in $t$ and we can use the chain rule on the last term.
\begin{align}f'(t)&=-\left(\frac{mg}{k} \right)+\frac{m^2g}{k^2}\frac{k}{m}\exp\left(-\frac{kt}{m}\right)\\&=-\left(\frac{mg}{k} \right)+\frac{mg}{k}\exp\left(-\frac{kt}{m}\right)\end{align}