The derivative of $1 - e^{-t/\tau}$ I am failing to understand how to compute the derivative of a few exponential functions.  Let's start with this one:
$$
v = 1 - e^{-t/\tau}
$$
The derivative is
$$
\frac{dv}{dt} = \frac{1-v}{\tau}
$$
Can someone walk me through this?  If this is explained somewhere else, I'd love to know where.
 A: $$\begin{align*}
\frac{d}{dt} (1 - e^{-t/\tau}) &= -\frac{d}{dt}e^{-t/\tau} \\
&= - (-1/\tau) e^{-t/\tau} \\
&= \frac{e^{-t/\tau}}{\tau} \\
&= \frac{1 - (1 - e^{-t/\tau})}{\tau} \\
&= \frac{1 - v}{\tau}
\end{align*}$$
A: We have $$\dfrac{dv}{dt} = \frac{1}{\tau} e^{-\frac{t}{\tau}} = \frac{1-(1-e^{-\frac{t}{\tau}})}{\tau}  = \frac{1-v}{\tau}$$. Is it clear?
A: $$\frac{dv}{dt}= \frac{d}{dt} (1 - e^{-t/\tau})$$
The derivative of $1$ is zero, so
$$= -\frac{d}{dt}e^{-t/\tau}$$
Using the chain rule,
$$= -e^{-t/\tau}\cdot\frac{d}{dt}\frac{-t}{\tau}$$
$$= -e^{-t/\tau}\cdot\frac{-1}{\tau}$$
$$= \frac{e^{-t/\tau}}{\tau}$$
Now, we know that $v=1 - e^{-t/\tau}$. Rearranging gives $e^{-t/\tau}=1-v$. Substitute above to get
$$= \frac{1-v}{\tau}$$
A: Derivative is a linear operator so it follows superposition:
$$\frac{d}{dt}(f(t)+g(t))=\frac{d f(t)}{dt}+\frac{d g(t)}{dt}$$
and
$$\frac{d}{dt}(k\times f(t))=k \times \frac{f(t)}{dt}~~~~~~~(k \text{ is constant number})$$
Also, derivative of some functions are known such as $\exp$, $\sin$ and $\cos$.
Putting all together, you can calculate derivative of this question.
In case a function derivative cannot be obtained by combinations of methods above, you need to use the general definition of derivation and solve a limit.
A: OK, let's walk:  first, note that
$v = 1 - e^{-t / \tau} \tag{1}$
yields, by direct differentiation, using the chain rule,
$\dfrac{dv}{dt} = -e^{-t / \tau} \dfrac{d(-t / \tau)}{dt} = \dfrac{1}{\tau} e^{-t / \tau}; \tag{2}$
from here on, it's simple algebra:
$\dfrac{1 - v}{\tau} = \dfrac{1 - (1 - e^{-t / \tau})}{\tau}$ $=\dfrac{1}{\tau} e^{-t / \tau} = \dfrac{dv}{dt}, \tag{3}$
using (2).  A pretty short walk, no?
