# Homework Help: Derivative of a unit step function

1. Mar 9, 2013

### Crabbz

So I'm busy struggling with some worked examples in my signals class. I understand the theory from the notes and textbook but I cannot seem to apply them to proper examples.
We are asked to find the derivative of g(t) = (1-e^(-t))*u(t) where u(t) is a unit step function.

I know the derivative of u(t) is the delta function, d(x). So when I try solve the derivative I use the chain rule and get:
g'(t) = e^(-t)*u(t) + (1-^e(-t))*d(x)
However I get stuck at this point and not sure where to go from here.

2. Mar 9, 2013

### vela

Staff Emeritus
That should be d(t), not d(x).

Use the fact that f(t)δ(t) = f(0)δ(t).

3. Mar 9, 2013

### Ray Vickson

You do not need the δ-function, since g(t) is continuous at t = 0 (no jump discontinuity). However, the left-hand and right-hand derivatives are different, so the answer will have a u(t) in it.

4. Mar 10, 2013

### tiny-tim

Welcome to PF!

Hi Crabbz! Welcome to PF!

(have a δ from the "Quick Symbols" box next to the Reply box )
(as vela says, of course that should be δ(t) )

Technically, it's not the delta function, it's the dirac delta, which isn't a function at all, it's a "generalised function", or "distribution", which really only works inside an integral.

(unfortunately, they're both written δ ! )

In particular, dirac δ(0) is not 1, it's actually ∞ (just look at the graph … how else can you get its total integral to equal 1 ?! )
Let's take an easier example, f(x) = (2x + 3)u(x) …

just by looking. that's the negative x-axis, then a line of slope 2 starting at (0,3): so its derivative is obviously 0 for x < 0, and 2 for x > 0, and there's a discontinuity at x = 0.

Now let's apply the chain rule, instead of just looking:

f'(x) = 2u(x) + (2x + 3)u'(x)

for x ≠ 0, that gives us f'(x) = 2u(x) + 0 (which is correct for x≠ 0)

for x = 0, it doesn't really tell us anything, since u'(0) = ∞​

The way to interpret f'(x) = 2u(x) + (2x + 3)u'(x) at x = 0 is to look at a small interval (h,k), with k > 0, and integrate:

∫ f'(x) dx = ∫ 2u(x) dx + ∫ (2x + 3)u'(x) dx

the LHS is f(k) - f(h), ie 2(k - h) if h ≥ 0, and 2k + 3 if h < 0

the first term on the RHS is 2(k - h) if h ≥ 0, but only 2k if h < 0

the second term on the RHS is (2*0 + 3)*0 = 0 if h ≥ 0, and (2*0 + 3)*1 = 3 if h < 0