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Derivative of a square root fraction.

  1. Jul 25, 2008 #1
    I need help finding the derivative of the following equation. This may look a little messy because it involves a square root and a fraction.

    y = square root of: 1 + x2 / 3 + 3 - x / 5

    My first thought is to change the equation to look like this:

    y = (1 + x2)1/2 / 3 + 3 - x / 5

    but I am not sure what the proper protocol is for finding a derivative of this kind of equation.

    An example is given of a similar question:
    y = square root of: 1 + x2 / 3 + 0.5 - x / 5

    where the derivative is:
    dy/dt = 1/6 (1 + x2)- 1/2 (2x) - 1/5
    but I can't figure out where the 1/6 comes from, or the final term, 1/5.

    Help please? :cry:
     
  2. jcsd
  3. Jul 25, 2008 #2

    Redbelly98

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    and
    Maybe others can understand these functions, but as for me it seems that some parentheses are missing here. The way I'm reading the first expression, for example, it would simplify to become

    y = square root of: (x2 / 3) + (1 + 3) - (x / 5)
    = square root of: (x2 / 3) - (x / 5) + 4

    But that can't be what you meant ...

    Edit added:
    Okay, from what is given for dy/dt, what we really have must be

    [square root of (1 + x2)] / 3 + 3 - x / 5
    and
    [square root of (1 + x2)] / 3 + 0.5 - x / 5

    At any rate, Hurkyl has given good advice here.
     
    Last edited: Jul 25, 2008
  4. Jul 25, 2008 #3

    Hurkyl

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    It's built entirely out of things you know how to differentiate (sums, differences, quotients, powers, constants), so I don't really understand why there's a problem. Where are you getting stuck?

    Incidentally, try plugging in x = 1 and, in painstaking detail, compute the value of that expression. (Show all 7 steps, one at a time! No skipping. You may use decimal approximations and/or a calculator. I think there are 7 of them; I might be off by 1 or 2)

    You do exactly the same steps in computing the derivative (possibly in reverse order, depending on how you actually think about such things) -- you just use the differentiation rules instead of the arithmetic rules.
     
    Last edited: Jul 25, 2008
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