# Antiderivative with chainrule

1. Aug 4, 2009

### caljuice

The text is giving me steps on how to do it but I don't get what's it is asking.

The equation is (v^-.5)v'=k where v= volume

Then says we consider the chain rule together with elementary antiderivative formulas to determine an antiderivative with respect to t of (v^-.5)v'

I haven't learned integrals yet. I'm exactly sure i'm even looking for. I don't really need an answer but explanation how chain rule can applies here. Since V is a function of t, do I expand v' to v'*dv/dt?

2. Aug 4, 2009

### iamthegelo

v' already is dv/dt, remember u(v(t))' = u'(v(t))*v'(t)

3. Aug 4, 2009

### caljuice

ah okay thanks. So then the antiderivative should be

2(V^.5)=kt+C ?

4. Aug 4, 2009

### rock.freak667

That would be correct. If you are ever unsure, just differentiate your answer and see if you get what you started with.