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Refresher on Derivatives

  1. Mar 7, 2007 #1
    [SOLVED] Refresher on Derivatives

    I feel rather dumb asking this but I could use a quick refresher on some parts of derivatives since for the past months I have been solely working on integrals and known cross sections so suddenly switching back has left my brain reeling.

    The main question I need explained is how do you take a derivative of [tex]-(t + 1)sin(\frac{t^2}{2})[/tex]

    I remember a trick my AP teacher told me of 1D2 + 2D1 meaning take the first part muliplied by the derivative of the second plus the second multiplied by the derivative of the first. The problem is, I don't remember how to take the derivatve of (t^2)/2
     
    Last edited: Mar 7, 2007
  2. jcsd
  3. Mar 8, 2007 #2
    Power rule. (t^n)'=nt^(n-1). Remember that constants slide out as well.
     
  4. Mar 8, 2007 #3
    You need to use a combination of the product rule and chain rule to differentiate the whole expression. But for t^2/2, use the power rule.

    For a quick refresher of the rules, you can have a look at this thread at the tutorial
    https://www.physicsforums.com/showthread.php?t=139690
     
  5. Mar 8, 2007 #4
    [tex]\frac{d}{dt}\left[-(t + 1)sin(\frac{t^2}{2})\right][/tex]

    [tex]= -\left[(t+1)\frac{d}{dt}\left(\sin \frac{t^2}{2}\right) + \left(\sin \frac{t^2}{2}\right)\frac{d}{dt}(t+1)\right][/tex]

    [tex]= -\left[(t+1)\cos \left(\frac{t^2}{2}\right)\frac{d}{dt}\left(\frac{t^2}{2}\right) + \sin \frac{t^2}{2}\right][/tex]

    [tex]= -\left[(t+1)\cos \left(\frac{t^2}{2}\right)\cdot t + \sin \frac{t^2}{2}\right][/tex]
     
    Last edited: Mar 8, 2007
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