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Antiderivative with chainrule

  1. Aug 4, 2009 #1
    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. jcsd
  3. Aug 4, 2009 #2
    v' already is dv/dt, remember u(v(t))' = u'(v(t))*v'(t)
  4. Aug 4, 2009 #3
    ah okay thanks. So then the antiderivative should be

    2(V^.5)=kt+C ?
  5. Aug 4, 2009 #4


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    Homework Helper

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