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The First Principles

  1. Apr 18, 2008 #1
    [SOLVED] The First Principles

    This is my question:

    1) Write down the definition of ƒ'(3), where [tex]f(x) = \sqrt{x + 1}[/tex]

    2) Use the definition to evaluate ƒ'(3).

    This isn't an exercise from a book etc.. so I don't know what's the correct solution to this question but here's my attempt:

    By the first principles( [tex]f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}[/tex] ):
    ∴ [tex]\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}[/tex]

    So would the definition of ƒ'(3) be like this:

    [tex]f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}[/tex]

    Simplifying gives:
    [tex]= 1[/tex]

    But another time I didn't use the first principles:

    [tex]f(x) = \sqrt{x + 1}[/tex]



    [tex]\frac{1}{1/2.\sqrt{3+1}} = 1[/tex]

    (?) - "?"

    I got the same answer using both methods (=1). Am I on the right track? What else do I need to show?

    Thank you. :smile:

    Last edited: Apr 18, 2008
  2. jcsd
  3. Apr 18, 2008 #2
    Please recheck what you get for this:

    [tex]f'(x) = \frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}[/tex]

    Remember you need to substitute x+h for x.
  4. Apr 18, 2008 #3
    You mean like:
    [tex]\frac{(x+h) - \sqrt{x+1}}{h}[/tex]

    [tex]\frac{(3+h) - \sqrt{3+1}}{h}[/tex]

    [tex]3 - 2[/tex] , → [tex] = 1[/tex]

    But is the rest of my above working correct? Did I show everything that was needed?
    What would be the right solution to my question? :rolleyes:
  5. Apr 18, 2008 #4

    Gib Z

    User Avatar
    Homework Helper

    No thats not what he meant :( Recheck your working, if f(x) = sqrt( x+1), what is f(x+h) ? Its not sqrt(x+h).
  6. Apr 18, 2008 #5


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    Staff Emeritus
    Science Advisor

    No, for two reasons (1) you forgot the "+1" in the first term and (2) you forgot the limit !!
    [tex]f'(x)= \lim_{h\rightarrow 0}\frac{\sqrt{3+h+1}-\sqrt{3+1}}{h}[/tex]

    No, it does not! Even forgetting the "+1" [itex]\sqrt{3+h}- \sqrt{3+1}[/itex] is NOT equal to h and so the fraction is not equal to 1!


    [tex]\frac{1}{2\sqrt{3+1}}= \frac{1}{4}[/tex]

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