# 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) ?

tiny-tim
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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
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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
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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..

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.