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At what rate is this function increasing?

  1. Apr 13, 2010 #1
    When x = 16,the rate at which [tex]\sqrt x[/tex] is increasing is [tex]\frac {1}{k}[/tex] times the rate at which x is increasing. What is the value of k?
  2. jcsd
  3. Apr 13, 2010 #2
    I thought it would be 4 but the answer is 8.
  4. Apr 13, 2010 #3
    Its the ratio of the derivatives evaluated at x=16,

    (2 sqrt(x))^-1
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