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Derivative problem

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data

    y=sqrt(x^2-1) find d^2y / dx^2

    2. Relevant equations

    power rule

    3. The attempt at a solution

    (x^2 - 1)^1/2
    y`=1/2(x^2 - 1)^-1/2
    y``= 1/2 * -1/2 (x^2 - 1)^-3/2
    y``= -1 / 4(x^2 - 1)^3/2

    Why does my book give the same answer without the 4 in the denominator? Doesn't find d^2y / dx^2 just mean find y``?
     
  2. jcsd
  3. Sep 13, 2007 #2
    you forgot the chain rule...

    y' should be = x(x^2 -1)^(-1/2)
     
  4. Sep 13, 2007 #3
    Ok, I have a few questions here.

    1. When do I know I have to use the chain rule?
    2. What does something like d^3y/dx^3 mean?
    3. Is this attempt correct?:


    y= (x^2 - 1)^1/2
    y`= = x(x^2 -1)^(-1/2)
    y``= x* -1/2 (x^2 - 1)^-3/2
    y``= -x / 2(x^2 - 1)^3/2
     
  5. Sep 13, 2007 #4

    Dick

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    1. If you know a rule to differentiate f(x) (like the power rule) and you see f(g(x)) where g(x) is more complicated than just x, you need to use the chain rule.
    2. It means y''' in the notation you are using.
    3. No. Now you are forgetting the product rule as well, and you are still not using the chain rule.
     
  6. Sep 13, 2007 #5
    1.Suppose for an example you try to take a derivative of something within a quantity that is powered

    [tex](x^2+2)^3[/tex]

    The whole function is cubed, and the only method to take the derivative of such function is to use the chain rule - take the derivative of outside function, then multiply the whole function by the inner derivative.

    In this case, the derivative of this function would be [tex](3(x^2+2)^2) * 2x[/tex]

    2.[tex]d^3y/dx^3[/tex] basically means to take the derivative of the function y with respect to the x, 3 times. In this case, it is equilivalent to y'''.

    [EDIT]
    3.No, your work is incorrect.
     
    Last edited: Sep 13, 2007
  7. Sep 13, 2007 #6

    Kurdt

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    1.) You use the chain rule when you differentiate a function of another function. In your example you have [itex] f(g(x)) = \sqrt{x^2 - 1} [/itex], where [itex] g(x) = x^2 -1 [/itex] and [itex] f(u) = \sqrt{u}, u = g(x) [/itex]. It just takes practise to recognise these things.

    2.) That would mean you differentiate the function three times. Any more than that description will require a mathematician.

    3.) You are fine after the first differentiation but after that you will need to apply the quotient rule.

    EDIT: Smeg! I type too slow.
     
  8. Sep 13, 2007 #7
    thank all of you for the information. My (foreign) TA is very hard to understand -- especially when I'm trying to learn calculus!

    I'm still confused, however. given
    y`= = x(x^2 -1)^(-1/2) -- which is supposedly correct, do I do a second differentiation with the power rule, or do I put (x^2 -1)^(-1/2) in the denominator to give it a positive power and then use the quotient rule?
     
  9. Sep 13, 2007 #8
    no, you can avoid using quotient rule on this, it won't be fun at all.

    Refer to using product rules, it'll simplify your life.
     
  10. Sep 13, 2007 #9

    Kurdt

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    Well both will give you the same answer. If you're feeling sadistic you can try them both to convince yourself.
     
  11. Sep 14, 2007 #10
    *sigh* i still manage to screw it up. I keep getting:

    1/2(x^2 - 1)^-1/2 * 2x
    x(x^2 - 1)^-1/2
    x / (x^2 - 1)^1/2

    what the heck am I doing wrong !?
     
  12. Sep 14, 2007 #11
    so what you have so far is

    [tex]y=(x^{2}-1)^{\frac{1}{2}}[/tex]

    [tex]y'=\frac{1}{2}(x^{2}-1)^{-\frac{1}{2}}\times2x[/tex]

    what would your next step be? don't use the rules of exponents to make your exponent positive, keep them all in the numerator.

    is your third line your 2nd derivative? i don't think it is, so you need to take the 2nd.

    do you notice how the 2's in the denominator will cancel, and the exponent of your x term at the front will increase if you continue to take the derivative?
     
    Last edited: Sep 14, 2007
  13. Sep 14, 2007 #12
    Ok let's see what I can do to help you.

    You have x(x^2)^-1/2

    Now let's do this.

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

    Notice the ' which means prime. So make sure you did the derivative correctly.

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

    Reduce reuse and recycle.
     
  14. Sep 14, 2007 #13

    Kurdt

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    Lets tidy this up before we proceed. You had arrived at this point:

    [tex] \frac{dy}{dx} = \frac{x}{\sqrt{x^2-1}} [/tex]

    Now you want to apply either the product rule or the quotient rule. I'll take you through the product rule first. The product rule is:

    [tex] \frac{d}{dx} f(x)g(x) = \frac{df(x)}{dx}g(x) + \frac{dg(x)}{dx}f(x) [/tex]

    If you choose [itex] f(x) = x [/itex] and [itex] g(x) = (x^2-1)^{-\frac{1}{2}} [/itex] can you proceed from there?

    As for the quotient rule:

    [tex] \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{\frac{df(x)}{dx}g(x)-\frac{dg(x)}{dx}f(x)}{(g(x))^2} [/tex]

    You can set [itex] f(x) = x[/itex] and [itex] g(x) = \sqrt{x^2-1} [/itex] and take it from there.

    If you can post the specific part you are having trouble with then we may have better luck directing you.
     
    Last edited: Sep 14, 2007
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