# Derivative of unit step

ok letsa say i have d/dt {(u(-2-t) + u(t-2)}

I know that d/dt { u(t) } is q(t)...

now is it correct to think that d/dt {(u(-2-t) + u(t-2)} = q(-2-t) + q(t-2) ?

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
HI ColdStart! ok letsa say i have d/dt {(u(-2-t) + u(t-2)}

I know that d/dt { u(t) } is q(t)...

now is it correct to think that d/dt {(u(-2-t) + u(t-2)} = q(-2-t) + q(t-2) ?

Nope … try again, using the chain rule (with g = -2-t) …

what do you get? ok so:
d/dt {(u(-2-t) + u(t-2)} = d/dt{ u(-(t+2)) + u(t-2)} = -q(t+2) + q(t-2) ?

tiny-tim
Science Advisor
Homework Helper
ok so:
d/dt {(u(-2-t) + u(t-2)} = d/dt{ u(-(t+2)) + u(t-2)} = -q(t+2) + q(t-2) ?

Yes , except it should still be -(t+2) inside the first q, shouldn't it? well then it turns out that its what i wrote in my first post, but in one book it shows -q(t+2)... thats why i got confused and was asking it here..

tiny-tim
Science Advisor
Homework Helper
well then it turns out that its what i wrote in my first post, but in one book it shows -q(t+2)... thats why i got confused and was asking it here..

No, in your first post you had …
q(-2-t) + q(t-2) ?

with no minus in front of the q.

(the book result would be the same if q is an odd function)

cool thanks!

Is $$\frac{du}{dt}=\delta (t)$$ valid for all t?

But the function u(t) is not continuous at t=0. 