1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiate this.

  1. May 4, 2014 #1

    adjacent

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Differentiate the function ##f(x)=\sqrt{x}##

    3. The attempt at a solution
    Power rule:
    $$\sqrt{x}=x^{\frac{1}{2}}$$
    $$\frac{d}{dx}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}}$$

    This is simple.However,I want to solve it using the definition of the derivative
    $$\lim_{\Delta x \to 0}=\frac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$$

    I don't know how I should deal with those surds. I can't simply subtract those two sqrts.
    :confused:
     
  2. jcsd
  3. May 4, 2014 #2

    CAF123

    User Avatar
    Gold Member

    Try multiplying numerator and denominator by ##\sqrt{x+\Delta x} + \sqrt{x}##.
     
  4. May 4, 2014 #3

    adjacent

    User Avatar
    Gold Member

    That will get cancelled and I will get the same thing again!
     
  5. May 4, 2014 #4

    CAF123

    User Avatar
    Gold Member

    Well yes, ofcourse since we are effectively multiplying by one. But the idea here is to multiply out the terms on the numerator and denominator. I'll start you off: $$\lim_{\Delta x \rightarrow 0}\, \frac{\sqrt{x+\Delta x} - \sqrt x}{\Delta x} \cdot \frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}} = \lim_{\Delta x \rightarrow 0}\, \frac{x + \Delta x - x}{\Delta x (\sqrt{x+\Delta x}+ \sqrt{x})}\,\, \dots $$
     
  6. May 4, 2014 #5

    adjacent

    User Avatar
    Gold Member

    Ok.I get $$\frac{\Delta x}{\Delta \sqrt{x+\Delta x}+ \Delta \sqrt{x}}$$

    How do I multiply a constant with it?
    I think it should be this: ##c . \sqrt{x}=\sqrt{c^2.x}## (As This is the same as ##\sqrt{c^2}.\sqrt{x}=c.\sqrt{x}##

    So after applying this rule,I get:
    $$\frac{\Delta x}{\sqrt{x.\Delta x +{\Delta x}^2}+\sqrt{x.{\Delta x}^2}}$$
    Then what? I can't get anywhere with this. :confused:
     
  7. May 4, 2014 #6

    CAF123

    User Avatar
    Gold Member

    I am not quite sure what you did there - remember ##\Delta x## is just a number so it obeys the usual distributivity, so ##\Delta x ( \sqrt{x + \Delta x} + \sqrt{x} ) = \Delta x \sqrt{x + \Delta x} + \Delta x\sqrt{x} ## if you wanted to multiply out. But notice this is not required - you are left with a ##\Delta x## on the numerator (as you got) and there is a ##\Delta x## on the denominator. So they cancel and what are you left with?
     
  8. May 4, 2014 #7

    adjacent

    User Avatar
    Gold Member

    Ohhhhhhh.
    Yeah.
    $$\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}=\frac{1}{\sqrt{\Delta x +x}+\sqrt{x}}$$
    $$\lim_{\Delta x \to 0}=\frac{1}{\sqrt{x+0}+\sqrt{x}}=\frac{1}{2.\sqrt{x}}$$


    Thank you sooo much.
    On question though,
    How did you know that multiplying by ##\frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}}## would make the weird surds go away?
    I have never really learned surds in school.So I don't know how it's working.
     
  9. May 4, 2014 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You don't want to take the "[itex]\Delta x[/itex]" inside the square root. Just cancel the [itex]\Delta x[/itex] outside the square root with the [itex]\Delta x[/itex] in the numerator. That leaves [itex]\frac{1}{\sqrt{x+ \Delta x}+ \sqrt{x}}[/itex]
     
  10. May 4, 2014 #9

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    It's a trick that you have to see once to know it. We essentially just apply the identity ##(a+b)(a-b) = a^2 - b^2##.

    Try to find out the derivative of ##f(x) = \sqrt[3]{x}## as an exercise using the identity ##a^3 - b^3 = (a-b)(a^2 + ab + b^2)##.

    Even better, try to find an identity for ##a^n-b^n## and try to find the derivative of ##f(x) = \sqrt[n]{x}##.
     
  11. May 4, 2014 #10

    CAF123

    User Avatar
    Gold Member

    I believe it is called 'rationalizing the numerator or denominator' or something like that. For example, if you have the number ##a/\sqrt{b}##, then you can rationalize the denominator (i.e make the surd go away) by simply multiplying by ##\sqrt{b}##. But if you do this, then you must also multiply the numerator by ##\sqrt{b}## and so in effect the net effect is that you are simply multiplying by one, so the number is unchanged. The result is then that ##a/\sqrt{b} = a\sqrt{b}/b##.

    In our case, I looked to rationalizing the numerator. ##\sqrt{x+\Delta x} + \sqrt{x}## does the job. But then I must also multiply by the same expression on the denominator too.
     
  12. May 4, 2014 #11

    adjacent

    User Avatar
    Gold Member

    Thanks.I think I have a lot to learn in A levels :smile:

    I will try them for sure.Thanks! :smile:
     
  13. May 4, 2014 #12

    Mark44

    Staff: Mentor

    Don't write ##\lim_{Δx \to ...}## by itself, as you did above and in your first post. That limit thing has to be attached to the function you're taking the limit of. It goes away when you actually take the limit.
    A standard trick when you have a sum or difference of radicals in either the numerator or denominator is to multiply by the conjugate (the same two terms, but separated by the opposite sign). The basic idea is that (a + b)(a - b) = a2 - b2. This can be useful for getting rid of square roots.
     
  14. May 4, 2014 #13

    Curious3141

    User Avatar
    Homework Helper

    Just for interest's sake, another way to arrive at the answer is to use the binomial/Taylor series, simplify then take the limit. You'll find that only the first order term is significant. But this way is less satisfying because the proof of the series expansion involves calculus in the first place. The elementary method presented here is far better.
     
  15. May 4, 2014 #14

    adjacent

    User Avatar
    Gold Member

    oops.That was a typo mistake.

    Binomial theorem!
    Is it difficult to learn? Can I learn it now?With knowing only differential calculus?
     
  16. May 4, 2014 #15

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    The binomial theorem is easy to learn and you can do it right now. It involves expressions of ##(a+b)^n## where ##n## is a positive integer.
    However, Curious3141 meant binomial series, which gives expressions of ##(a+b)^n## where ##n## is much more general. You need to know Taylor series for this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Differentiate this.
  1. Differentiate This (Replies: 9)

  2. Differentiating this (Replies: 7)

  3. Differentiable ? (Replies: 3)

  4. Differentiable (Replies: 12)

Loading...