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Find the derivative from first principles

  1. Jun 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the derivative from first principles, i.e, from the definition of f ' as a limit, for f(x)= [tex]\sqrt{}a-bx, where a and b are positive constants. What are the domains of f and f ' ?

    3. The attempt at a solution
    I solved:

    f(x) = \sqrt{a-bx} ---> f(x+h) = \sqrt{a-b(x+h)}

    lim h-->0 = \sqrt{a-b(x+h)} - \sqrt{a-bx} / h

    lim h-->0 = \sqrt{a-bx+bh-a+bx} / h

    lim h-->0 = \sqrt{bh} / h

    lim h-->0 = \sqrt{bh}* \sqrt{bh} / h\sqrt{bh}

    lim h-->0 = b/ \sqrt{bh}

    What did I do wrong here?

    Please help and thanks
    Last edited: Jun 3, 2008
  2. jcsd
  3. Jun 3, 2008 #2


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    Homework Helper

    Check this step. The former does not lead to the latter. You cannot "extend" the root from the left expression of the numerator to the right one. Instead, multiply the numerator (and the corresponding denominator) by its conjugate; ie. Conjugate of a+b is a-b and vice versa.
  4. Jun 3, 2008 #3
    hmmm ....

    ok then i should do lim f(x) - g(x) = lim f(x) - lim g(x)

    = lim sf(a-b(x+h)) * sf(a-b(x+h)/ h sf(a-b(x+h)) * lim sf(a-bx) * sf(a-bx)/ h sf(a-bx)

    = lim a-bx+bh / h sf(a-bx+bh) * lim (a-bx) / h sf(a-bx)

    I dont know how to cancel h
  5. Jun 3, 2008 #4
    \frac{\sqrt{a-b(x+h)} - \sqrt{a-bx}}{h}

    Try multiplying the numerator by
    \frac{\sqrt{a-b(x+h)} + \sqrt{a-bx}}{\sqrt{a-b(x+h)} + \sqrt{a-bx}}

    too hard to read what you wrote... >>
  6. Jun 3, 2008 #5
    too hard to read what you wrote... >>[/QUOTE]

    Sorry, i dont know to use those symbols. Would you show me please?
  7. Jun 3, 2008 #6
    Basically, you multiply by the conjugate to remove the radicals in this situation.

    For example, [tex](\sqrt{c}+\sqrt{d})(\sqrt{c}-\sqrt{d})=c-d[/tex]. where [tex](\sqrt{c}-\sqrt{d})[/tex] is the conjugate of [tex](\sqrt{c}+\sqrt{d})[/tex]
  8. Jun 3, 2008 #7
    How do I use square roof symbol please?
  9. Jun 3, 2008 #8
    \sqrt{} you have to wrap it around "tex" and its end tag "/tex"
  10. Jun 3, 2008 #9
    That how I did but when I posted it did not show square roof's symbol.
    Probably work's computer doesnt allow me to do it.
  11. Jun 3, 2008 #10


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    Staff Emeritus
    Science Advisor
    Gold Member

    Here is the thread that introduced latex:


    If you click on any latex image you can see the code used. There are a couple of websites where you can practise your code linked to at the end of that thread. The forum has a drop down menu that looks like a capital sigma for adding things in latex.
  12. Jun 3, 2008 #11
    is this the right answer?

    - b
    sf (a-bx) + sf (a-bx)

    so the domain of f ' is a-bx > 0
  13. Jun 3, 2008 #12


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    Staff Emeritus
    Science Advisor

    "sf" is square root?

    Yes, the derivative of [itex]\sqrt{a-bx}[/itex] is
    just what you have.
  14. Jun 3, 2008 #13
    Solve for the domain of f, and f' in terms of x. Btw, it's a good thing to know that it is square root and not square "roof" =P
  15. Jun 3, 2008 #14
    my bad.... hehehehe ( I've been here just couple months)
    Thanks so much all.
  16. Jun 3, 2008 #15
  17. Jun 3, 2008 #16
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