# Derivative of unit step function?

How would I find the derivative of a unit step function? I understand that the unit impulse function will be used but I'm not sure how to use it.

I am trying to find the derivative of this:

$v(t) = u(t+1) - 2u(t) + u(t-1)$

$u(t) = 0$ when $t < 0$

$u(t) = 1$ when $t > 0$

The relationship between unit step function and impulse function:

δ(n) = u(n) - u(n-1)

$δ(t)=du(t)/dt$

• Continuous $\displaystyle \delta(t) = \frac{d u(t)}{dt}$
– user186104
Nov 1, 2016 at 0:15
• Discrete $\displaystyle \delta[n] = u[n] - u[n-1]$
– user186104
Nov 1, 2016 at 0:17
• Possible duplicate of How to prove that the derivative of Heaviside's unit step function is the Dirac delta?.
– user137731
Nov 1, 2016 at 0:22
• @arthur Thanks for the clarification. I've edited the post.
– zdub
Nov 1, 2016 at 0:24
• @zdub did you mean for one of the Heaviside functions to be $u(t\color{red}{+}1)$? Otherwise the first and third terms are the same and can be simplified to $2u(t-1)$.
– user137731
Nov 1, 2016 at 0:26

The derivative of unit step $$u(t)$$ is Dirac delta function $$\delta(t)$$, since an alternative definition of the unit step is using integration of $$\delta(t)$$ here.

$$u(t) = \int_{-\infty}^{t} \delta(\tau) d\tau$$

Hence, $$\frac{dv}{dt} = \delta(t+1) - 2\delta(t) + \delta(t-1)$$

• Thanks! So the graph of the derivative would just be vertical lines at t = -1, 0 and 1?
– zdub
Nov 1, 2016 at 0:21
• @zdub The way that engineers draw it, there should be vertical arrows at $1$ and $0$, each of which is $2$ units tall (but at $0$ it should point in the opposite direction).
– user137731
Nov 1, 2016 at 0:24
• @zdub You have two $u(t-1)$ in your question, or $2u(t-1)$. This means the derivative is a $2\delta(t-1)$ that is represented by an upward arrow with amplitude $2$ at $t=+1$. If your actual question is $u(t-1)+2u(t)+u(t+1)$, then there will be three upward delta functions in the derivative: two of them are at $t=\pm1$ with unit scale, and the third at origin with scale $2$.
– msm
Nov 1, 2016 at 0:33
• Just to clarify: Dirac delta is NOT a function. See math.stackexchange.com/questions/285642/… Oct 16, 2022 at 6:07