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Homework Help: What is the derivative of x.(rootx) from first principles

  1. Sep 23, 2005 #1
    The title is pretty self explanitary.

    From first principles find f'(x) for f(x)=x.(rootx)

    Look forward to your replies
  2. jcsd
  3. Sep 23, 2005 #2

    Tom Mattson

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    *sniff sniff*

    This smells a lot like homework to me!

    If you want to calculate any derivative from first principles, you need to go back to the definition of the derivative (you can't get more first principled than that). If you recognize that [itex]x\sqrt{x}=x^{3/2}[/itex], you should be able to do it with a minimum of trouble.

    Give it a shot and see if you can't do it.
  4. Sep 23, 2005 #3
    f'(x) = f(a+h) + f(a)

    f'(x) = lim (h->0) [(x+h)^3/2 + x^3/2]/h

    how do you resolve the (x+h)^3/2)

  5. Sep 23, 2005 #4


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    Well first of all that's wrong: f'(x) is not f(a+h)- f(a).

    In your second line, you at least have the limit but it's still wrong- remember the words difference quotient. You need a "difference" (so it is f(x+h)- f(x) not "+") and you need a "quotient"- divide by h.

    [tex]f'(x)= lim_{h->0}\frac{(x+h)^{\frac{3}{2}}-x^{\frac{3}{2}}}{h}[/tex]

    Try "rationalizing the numerator"- multiply both numerator and denominator by
    [tex](x+h)^{\frac{3}{2}}+ x^{\frac{3}{2}}[/tex]
  6. Sep 23, 2005 #5

    Tom Mattson

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    This could not possibly have come from a textbook. The definition of the derivative of a function is

    [tex]f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex].

    Almost. The numerator should be a difference (not a sum).

    You rewrite the numerator correctly, and then you multiply the top and bottom of the difference quotient by the conjuagate of the numerator.

    This thread really belongs in the Homework section, so I'm going to move it there.
    Last edited: Sep 23, 2005
  7. Sep 24, 2005 #6
    I typed + instead of minus!!! The IT system was going down in 5 minutes so i rushed!

    I get to (after rationalising the numerator) to

    [tex]f'(x)=\lim_{h\rightarrow 0}\frac{(x+h)^3-x^3}{h[(x+h)^\frac{3}{2} + x^\frac{3}{2}]}[/tex].

    [tex]f'(x)=\lim_{h\rightarrow 0}\frac{3x^2h+3h^2x +h^2}{h[(x+h)^\frac{3}{2} + x^\frac{3}{2}]}[/tex].

    At this point i think I must have made an error for if the limit is taken you get 0/0 (all terms are multiplied by h).

    Any idea?
    Last edited: Sep 24, 2005
  8. Sep 24, 2005 #7


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    Just a small error, it's 3x2h + 3xh2 + h3 in the numerator (not h2).
    And in the denominator it's x3 / 2, not 33 / 2
    3x2h + 3xh2 + h3 = h(3x2 + 3xh + h2)
    Note that the numerator and denominator have 'h' in common, so what can you do next?
    Viet Dao,
    Last edited: Sep 24, 2005
  9. Sep 24, 2005 #8
    My error came because i had divided by h in my next line ofworking on paper but thought i may have gone wrong! dividing by h gives

    [tex]f'(x)=\lim_{h\rightarrow 0}\frac{3x^2+3hx +h^2}{(x+h)^\frac{3}{2} + x^\frac{3}{2}}[/tex].

    If i take the limit here I get



    Last edited: Sep 24, 2005
  10. Sep 24, 2005 #9


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    Yup, now take the limit as h tends to 0. It's no longer 0 / 0.
    [tex]\lim_{h \rightarrow 0} \frac{3x ^ 2 + 3xh + h ^ 2}{(x + h) ^ \frac{3}{2} + x ^ \frac{3}{2}} = \frac{3x ^ 2 + 3x0 + 0 ^ 2}{(x + 0) ^ \frac{3}{2} + x ^ \frac{3}{2}}[/tex].
    Can you go from here??
    Viet Dao,
  11. Sep 24, 2005 #10
    Look above I realisd I was being stupid!
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