Derivative of $e^{- t}$

Find $\frac{d}{dt} \left(e^{- t}\right)$.

The function $e^{- t}$ is the composition $f{\left(g{\left(t \right)} \right)}$ of two functions $f{\left(u \right)} = e^{u}$ and $g{\left(t \right)} = - t$.

Apply the chain rule $\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$:

The derivative of the exponential is $\frac{d}{du} \left(e^{u}\right) = e^{u}$:

Return to the old variable:

Apply the constant multiple rule $\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$ with $c = -1$ and $f{\left(t \right)} = t$:

Apply the power rule $\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dt} \left(t\right) = 1$:

Thus, $\frac{d}{dt} \left(e^{- t}\right) = - e^{- t}$.

$\frac{d}{dt} \left(e^{- t}\right) = - e^{- t}$A