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Quick derivative question

  1. Sep 16, 2007 #1
    Where does this 2x (highlighted) come from? I thought this step was just the 'product rule', but it looks like the 'chain rule' was applied as (x^2 + 2)`. Were these two rules used simultaneously?

  2. jcsd
  3. Sep 16, 2007 #2


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    First they used the product rule, and then they used the chain rule to find the derivative of sqrt(x^2+2). The 2x comes from the derivative of x^2+2.
  4. Sep 17, 2007 #3


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    That is actually a double application of the chain rule. To differentiate [itex]\sqrt{x^2+1}[/itex], let u= x2 and let v= u+ 1. [itex]\sqrt{x^2+ 1}[/itex] becomes [itex]\sqrt{v}= v^{1/2}[/itex]. It's derivative is (1/2)v-1/2v'. Of course, v'= u'= 2x.

    (I am ignoring the x multipying the square root since you only asked about the chain rule.)
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