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Homework Help: Help with derivative

  1. Jul 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Given the equation f(x) = 2x+4/√x evaluate:

    (1) f(0.5)
    (2) f '(0.5)

    2. Relevant equations

    3. The attempt at a solution

    (1) is simply to substitute the value 0.5 into the equation

    (2) however i am assuming is to find the derivative of the equation

    This is where i get stuck. Its so basic i know but i always seem to struggle with the simple things.

    this is where im at

    f(x) = 2x+4/√x
    = 2x/√x + 4/√x

    The reciprocal of√x is x^1/2 so im assuming the 2x/√x is 2x(x)^-0.5 and this is how far i get on this part of the equation.

    The 4/√x im sure = 4x^-0.5 and using the 'direct rule' becomes:-

    (-0.5)(4)x^-0.5 - 1

    converting to root from:-


    from there i get lost.

    I would greatly appreciate a step by step solution.

    Thanks guys
  2. jcsd
  3. Jul 17, 2012 #2


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    You wrote f(x) = 2x + 4/sqrt(x), which would equal

    [tex]f(x) = 2x + \frac{4}{\sqrt{x}}[/tex]

    under the standard rules of operator precedence. However, based on what you subsequently wrote, I am guessing the function is actually

    [tex]f(x) = \frac{2x + 4}{\sqrt{x}}[/tex]

    which, if you wanted to write it "in line", would be f(x) = (2x + 4)/sqrt(x) [note the parentheses]. This can be simplified as

    [tex]f(x) = \frac{2x}{\sqrt{x}} + \frac{4}{\sqrt{x}} = 2\sqrt{x} + \frac{4}{\sqrt{x}}[/tex]

    Now, what is this in terms of powers of x?
  4. Jul 17, 2012 #3
    ill have a go

    2x^0.5 + 4x^-0.5

    am i on the right path?
  5. Jul 17, 2012 #4


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    Yes, that's right. So what do you get if you differentiate this?
  6. Jul 17, 2012 #5


    Staff: Mentor

    As others have noted already, the right side needs to be written as (2x + 4)/√x.
    No, the reciprocal of √x is NOT "x^1/2". The reciprocal of √x is 1/√x, or x^(-1/2). √x and x^(1/2) - note the parentheses) are two ways of writing the same thing.
    Please use parentheses around your exponents, like so:
    (-0.5)(4)x^(-0.5 - 1)
    and here, too.

    That's not the way it works here at Physics Forums. The rules don't permit us to provide you with a step-by-step solution, but we're happy to guide you as you do the work.
  7. Jul 17, 2012 #6
    Thanks for the reply guys.

    Totally understand Mark, thanks for your input:)

    jbunniii can you go back 1 step. You simplified 2x/√x to be 2√x. Can i ask how you got this?

    on a side note i am doing a self study course so i am my own tutor at present. I really want to get a handle on this so please bare with me.

  8. Jul 17, 2012 #7


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    If jbunniii will forgive me for jumping in here there are two ways to do that:
    first, note that [itex]x= (\sqrt{x})^2= (\sqrt{x})(\sqrt{x})[/itex] so that
    [tex]\frac{x}{\sqrt{x}}= \frac{(\sqrt{x})(\sqrt{x})}{\sqrt{x}}= \sqrt{x}[/tex]
    [tex]x/\sqrt{x}= x^1/x^{1/2}= x^1x^{-1/2}= x^{1- 1/2}= x^{1/2}= \sqrt{x}[/tex]

    Oh, dear. Another person who wants us to "bare" with him. This forum is getting too wild for me!
  9. Jul 17, 2012 #8
    Oh boy..........."bare".

    Is it too late to change that to "bear" :)
  10. Jul 18, 2012 #9
    Right where were we

    2x^(0.5) + 4x^(-0.5)

    = (1/2)(2)x^(0.5 -1) + (-1/2)(4)x^(-1/2 - 1)

    = x^(-1/2) + -2x^(-2/3)

    Am i right so far
  11. Jul 18, 2012 #10


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    Almost. What is -1/2 - 1? (Hint: it's not -2/3.)
  12. Jul 18, 2012 #11
    Hmmmmm now i am stumped.I think this is where im going wrong. Going by the calculator it is -2/3. Is it 2/3( -and- = + thus 1/2 + 1)

    i feel like a plonker:(
  13. Jul 18, 2012 #12


    Staff: Mentor

    Better get another calculator...

    If you have 1/2 pizza and 1 pizza, that's not the same as 2/3 pizza.
  14. Jul 18, 2012 #13
    wow it really has been a long day.

    ok ok. Round the wrong way.

    so its -3/2
  15. Jul 18, 2012 #14
    with the correction:-

    x^(-1/2) + -2x^(-3/2)

    x^(-1/2) = 1/√x

    2x^(-3/2) this is where im lost..................

    Thanks for the guidance guys
  16. Jul 18, 2012 #15


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    Well, x^(-1/2) - 2x^(-3/2) is a perfectly valid answer. If you don't like negative roots, then use the fact that

    [tex]x^{-t} = \frac{1}{x^t}[/tex]

    [edit] I just re-read the original post and recalled that it asks for f'(0.5). So you need to plug in x = 0.5, and simplify if desired.
  17. Jul 18, 2012 #16
    So just to finish off, removing the negative roots would =

    1/√x - 2/x^(3/2) ?
  18. Jul 18, 2012 #17


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    Looks right to me. Now you need to evaluate this at x = 1/2.
  19. Jul 18, 2012 #18
    Thank you SO much for your help.

    What a great forum :)
  20. Jul 18, 2012 #19
    Before we close this chapter, can i ask what is the "name" given when removing the negative roots?
  21. Jul 18, 2012 #20


    Staff: Mentor

    Far as I know, there's no name. What jbunniii showed is a property of exponents.
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