The derivative of $1 - e^{-t/\tau}$

Question

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.

Answer 1

$$\begin{align*} \frac{d}{dt} (1 - e^{-t/\tau}) &= -\frac{d}{dt}e^{-t/\tau} \\ &= - (-1/\tau) e^{-t/\tau} \\ &= \frac{e^{-t/\tau}}{\tau} \\ &= \frac{1 - (1 - e^{-t/\tau})}{\tau} \\ &= \frac{1 - v}{\tau} \end{align*}$$

Answer 2

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?

Answer 3

$$\frac{dv}{dt}= \frac{d}{dt} (1 - e^{-t/\tau})$$

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

$$= \frac{1-v}{\tau}$$

Answer 4

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.

Answer 5

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?