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State the domain of the function and the domain of its derivative

  1. Jun 23, 2008 #1
    Find the derivative. State the domain of the function and the domain of its derivative.

    f(x) = x + √x

    f(x) = (3 + x) / 1-3x

    Find F'(a)

    f(x) = (x^2 + 1) / (x - 2)

    f(x) = √3x + 1
     
  2. jcsd
  3. Jun 23, 2008 #2

    Defennder

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    And your work so far...?
     
  4. Jun 24, 2008 #3
    Re: questions

    For the first one i got up to x + √x = x^1/2 = x^3/2

    2nd one 1(1-3x) - (-3)(3 + x) all over (1-3x)^2 = 10 / (1 - 3x) ^2

    3rd one i know its the quotient rule but not sure where to go with this one

    For the fourth i tried the chain rule

    √3x + 1 = 3x^1/2 + 1 = 1/2(3x)^1/2
     
  5. Jun 24, 2008 #4

    HallsofIvy

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    Re: questions

    ? No, [itex]x+ \sqrt{x}[/itex] is NOT equal to [itex]x^{1/2}= \sqrt{x}[/itex] and NEITHER of those is equal to [itex]x^{1/2}[/itex]. Did you mean that [itex]\sqrt{x}= x^{1/2}[/itex]? And that the derivative is [itex]x^{3/2}[/itex]? That last is not correct, either. Surely, you know what the derivative of x= x1 is 1? And what is the derivative of [itex]\sqrt{x}= x^{1/2}[/itex]?

    Yes, that is correct!

    USE the quotient rule of course! What is (x2+ 1)' ? What is (x- 2)'? Do it just like you did number 2.

    Is that √(3)x+ 1, √(3x)+ 1, or √(3x+1)? In any case, none if those is equal to 3x^(1/2)+ 1.

    The derivative of √(3) x+ 1 should be trivial. √(3x)+ 1 can be done as √(3)x^(1/2)+ 1, and √(3x+1) should be done using the chain rule: √(3x+1)= √u with u= 3x+1:
    (√(3x+1))'= (du^(1/2)/du)(d(3x+1)/dx).
     
  6. Jun 24, 2008 #5
    Re: questions

     
  7. Jun 24, 2008 #6

    Defennder

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    Re: questions

    Use the power rule for differentiation. What is the derivative of x^n? Just apply it to x^1/2.
     
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