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Basic differentiation?

  1. Dec 4, 2003 #1
    Ok, In my calc book I am having problem with the first example they give in a chapter on Differentiation.

    It gives the equation of [itex]V = \frac{4}{3} \pi r^3[/itex] and then it is differentiated by time = t.

    [tex]\frac{dV}{dt} = \frac {dV}{dr} \frac{dr}{dt} = 4 \pi r^2 \frac {dr}{dt}[/tex]

    I dont quite understand why the chain rule is being used in the first place, because [itex]4 \pi r^2[/itex] doesn't need it, and also I dont understand where [itex]\frac{dr}{dt}[/itex]comes from in the last part. Im quite confused by this basic stuff [b(] so thanks for any help.

    edit : \/ Thanks for the help getting it to show up right :)
    Last edited: Dec 4, 2003
  2. jcsd
  3. Dec 4, 2003 #2


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    Here's the fixed text for you.
  4. Dec 4, 2003 #3
    Your calc book seems to be implying that [itex]r[/itex] is a function of [itex]t[/itex]. So when you differentiate with respect to [itex]t[/itex] you have no choice but to use the Chain Rule.

    Now clearly using the power rule we have:

    \frac{dV}{dr}=4\pi r^2

    But we need [itex]\frac{dV}{dt}[/itex], so we invoke the chain rule:

    \frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}=4\pi r^2\cdot\frac{dr}{dt}

    We know what [itex]\frac{dV}{dr}[/itex] is so we can substitute it into the equation, but we don't know what [itex]\frac{dr}{dt}[/itex] is so it remains.
  5. Dec 4, 2003 #4
    the chain rule was just a mathematical trick to cancel out the "dr"'s when using 2 fractions or more.

    but mathematical tricks work out great.
  6. Dec 4, 2003 #5
    Thanks alot, I am still having a problem understanding why exactly the chain rule is used. on the left side all you had to do was take the derivative of V and the derivative of t. I understand that [itex]\frac{dr}{dt}[/itex] is what I am trying to find, but I dont understand the reasoning behind why the chain rule is used to turn the right hand side into a derivative of time? Im a bit confused.
  7. Dec 4, 2003 #6
    On the left side, you aren't taking the derivative of V and the derivative of t. You're taking the derivative of V with respect to t. Basically, [itex]\frac{dV}{dt}[/itex] is a function that tells you how V changes when you change t.

    On the right side, you also take the derivative with respect to t. If you were taking the derivative with respect to r, it would be easy because you can just use the power rule. But you can't, because you need the derivative with respect to t.

    That's were the chain rule comes in. It lets you find the derivative of [itex]\frac{4}{3}\pi r^3[/itex] by first finding the derivative with respect to r, then multiplying that by the derivative of r with respect to t.
    Last edited: Dec 4, 2003
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