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Homework Help: Implicit Differentiation

  1. Oct 25, 2005 #1
    Can someone check my answer (I am trying to find the second derivative) for any mistakes?
    I have looked it over many times, and I've realized that my second derivative is not correct, but I cannot figure out why. Thank you.
    [tex]\sqrt{x} + \sqrt{y} = 1[/tex]
    [tex]\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}(y') = 0[/tex]
    [tex]y' = (\frac{-1}{2\sqrt{x}})(\frac{2\sqrt{y}}{1}) = \frac{-2\sqrt{y}}{2\sqrt{x}} = \frac{-\sqrt{y}}{\sqrt{x}}[/tex]
    [tex]y'' = \frac{(\frac{-y'}{2\sqrt{y}})(\sqrt{x}) + (\sqrt{y})(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2}[/tex]
    [tex]y'' = \frac{\frac{-y'\sqrt{x}}{2\sqrt{y}} + \frac{\sqrt{y}}{2\sqrt{x}}}{x}[/tex]
    [tex]y'' = \frac{-y'\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
    [tex]y'' = \frac{-(\frac{-\sqrt{y}}{\sqrt{x}})\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
    [tex]y'' = \frac{\sqrt{y}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
    [tex]y'' = (\frac{\sqrt{x}\sqrt{y}+y}{2x\sqrt{x}\sqrt{y}})(\frac{\sqrt{x}\sqrt{y}}{\sqrt{x}\sqrt{y}})[/tex]
    [tex]y'' = \frac{xy + y\sqrt{x}\sqrt{y}}{2x^2y}[/tex]
    [tex]y'' = \frac{x + \sqrt{x}\sqrt{y}}{2x^2}[/tex]
     
    Last edited by a moderator: Oct 25, 2005
  2. jcsd
  3. Oct 25, 2005 #2

    TD

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    Your first derivative is correct, but I don't see what you mean with 0y' then... You also appear to have a lot of equal-signs, it seems a bit weird...

    So we have:
    [tex]\left( {\sqrt x + \sqrt y } \right)^\prime = \frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}[/tex]

    Now, the second derivative is doing just the same, just a bit longer:
    [tex]\left( {\frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}} \right)^\prime = \left( {\frac{1}{{2\sqrt x }}} \right)^\prime + \left( {\frac{{y'}}{{2\sqrt y }}} \right)^\prime = \cdots [/tex]
     
  4. Oct 25, 2005 #3
    I don't understand your first function. Is that supposed to be the second derivative?
     
  5. Oct 25, 2005 #4
    And I have a lot of equal signs because I'm trying to simplify my second derivative. I think that's where I made the mistake... in simplifying.
     
  6. Oct 25, 2005 #5

    TD

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    My first line is the first derivative, the second line is the beginning of the second derivative. Since it's linear, I already split it for you. The first term will be easy since it only depends on x, the second will be a bit longer.
     
  7. Oct 25, 2005 #6
    Well that's what I did, starting in my fourth line. I used the quotient rule on my simplified form of the first derivative, to find the second derivative. Shouldn't both methods result in the same answer?
     
  8. Oct 25, 2005 #7

    TD

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    Ok, I wasn't aware of the fact it was an equation at first.
    It's much clearer now you've adjusted your initial post.

    Line 6 seems to be correct, then:

    [tex]y'' = \frac{{ - y'\sqrt x \sqrt x + \sqrt y \sqrt y }}{{2x\sqrt x \sqrt y }} = \frac{{y - y'x}}{{2x^{3/2} \sqrt y }}[/tex]
     
  9. Oct 25, 2005 #8
    How about after line 6?
    Is the whole thing correct? Right up to the last line?
     
  10. Oct 25, 2005 #9

    TD

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    I took over after line 6 but I didn't realise you substituted y' again.
    Anyway, it seems to correct (at least your solution, I didn't carefully check every step). It's possible to 'simplify' it one last bit, although it doesn't really matter a lot anymore:

    [tex]y'' = \frac{{x + \sqrt x \sqrt y }}{{2x^2 }} = \frac{{\sqrt x + \sqrt y }}{{2x^{3/2} }}[/tex]
     
  11. Oct 25, 2005 #10
    This makes sense, however...

    When I substitute a value of x, y, & y' into my equation on line 4 (the first y''), I get a different value for y'' than when I substitute the exact same values into my last equation for y''.

    Why is that?
     
  12. Oct 25, 2005 #11

    TD

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    You realise that in line 4, the first derivative y' is still present in the expression for y''?
     
  13. Oct 25, 2005 #12
    nevermind..
     
  14. Oct 25, 2005 #13
    Yes I know.
     
  15. Oct 25, 2005 #14
    I used my values as:
    x = 4
    y = 1
    y' = 1/2
     
  16. Oct 25, 2005 #15

    TD

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    Huh? Your initial equation was [itex]\sqrt x + \sqrt y = 1[/itex] but [itex]\sqrt 4 + \sqrt 1 \ne 1[/tex]...
     
  17. Oct 25, 2005 #16
    You are right.

    I know why now...
    I rearranged that formula to put it in the graphing calculator.
    But I forgot to consider the fact that squares can be positive or negative. ................ Well at least I think that's my mistake.
    Do you know what the graph of this function looks like?
    Is it a circle? Or not?
     
  18. Oct 25, 2005 #17
    Never mind that......

    Substitute x = 4 into the original equation and solve for y.

    You will get the value "1" .
    That's what I just got.

    This is odd.....
     
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