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Homework Help: Implicit Differentiation Clarification

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] 2x^{2} - 3y^{2} = 4 [/tex]

    2. Relevant equations
    We say that y is an implicit function of x if we are given an equation:

    [tex] \sigma(x,y) = \tau(x,y)[/tex]

    Then to differentiate we do:

    [tex] \frac {d(\sigma(x,y))} {dx} = \frac {d(\tau(x,y))} {dx} [/tex]
    3. The attempt at a solution

    [tex] 2x^{2} - 3y^{2} = 4 [/tex]

    [tex] \frac {d(2x^{2} - 3y^{2})} {dx} = \frac {d(4)} {dx} [/tex]

    [tex] 4x - 6y \frac {dy} {dx} = 0 [/tex]

    [tex] -6y \frac {dy} {dx} = -4x [/tex]

    [tex] \frac {dy} {dx} = \frac {4x} {6y} = \frac {2x} {3y} [/tex]

    If this is correct, that's great, but I really do not understand why the [tex] \frac {dy} {dx} [/tex] gets put into the equation. Of course I understand its function but I really don't understand the rules that go along with placing it in the equation. If I add it to one side shouldn't I have to add it to the other?

    If I am wrong, please tell me what I did wrong and how to fix it.

    Thanks!!
     
  2. jcsd
  3. Feb 23, 2010 #2

    Dick

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    That is great. Sure, dy/dx is what you want to find, isn't it? I don't see why you think you 'added it to one side'. You just differentiated and solved for it. Didn't you? There's nothing odd or wrong about that.
     
  4. Feb 23, 2010 #3
    Well, I guess I mean after this:

    [tex] \frac {d(2x^{2} - 3y^{2})} {dx} = \frac {d(4)} {dx} [/tex]

    you get:

    [tex] 4x - 6y \frac {dy} {dx} = 0 [/tex]

    Where does the dy/dx come from? It's not leftover from the equation on the left side so I must have added it to the equation, and as far as I know, when you add (or subtract or anything) to/from one side you must also do it to the other. So that's where I am second guessing myself I suppose because it feels wrong to me to add it to the left and not the whole equation.... if you get what I mean...
     
  5. Feb 23, 2010 #4

    Dick

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    You didn't 'add it'. You differentiated d(y^2)/dx and got 2*y*dy/dx using the chain rule. Or I at least sincerely hope you did. I'm now guessing you didn't. But that's what you should have done.
     
  6. Feb 23, 2010 #5
    um yeah I didn't, I mean I 'kind of' did, I took the derivative of [tex] 2x^{2}-3y^{2} [/tex] and then added dy/dx to the result because that's how it was explained in my book (note it didn't say to "add dy/dx" I just got that from the examples.) If possible could you further explain what you just said?
     
  7. Feb 23, 2010 #6

    Dick

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    Chain rule, man. f(y(x))'=f'(y(x))*y'(x). Look it up. f here is the squaring function. f(u)=u^2. So f'(u)=2u. f'(y(x))=2*y(x). y'(x)=dy/dx. Starting to make sense?
     
  8. Feb 23, 2010 #7
    Yeah, I guess I just never carried the thought process that far.... but I am pretty sure I understand you now. In my equation x is the independent variable and y depends on x so y is actually y(x), am I thinking about this correctly now?
     
  9. Feb 24, 2010 #8

    Dick

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    Right. So ((y(x))^2)'=2*y(x)*y'(x) where "'" is d/dx.
     
  10. Feb 24, 2010 #9
    Hey, whenever you get online can you help me out with this question Dick? It's the same type of question from the same section, I think I am just a little confused:

    [tex] y = \frac {1} {(x+y)} [/tex]

    [tex] \frac {d(y)} {dx} = \frac {d(\frac {1} {(x+y)})} {dx} [/tex]

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

    At this point I would consider myself stuck, the only move I see is this:

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

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

    and then you get 0 = which is not what I am trying to solve for obviously.. so I made a mistake some where, also I am not sure if

    [tex] 1 + \frac {dy} {dx} [/tex]

    in

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

    should be considered a binomial such as:

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

    Hopefully thats not all too confusing..., thanks for the help in advance!

    ______________________________________________________________________

    Another question just to check to see if I understand this:

    [tex] y^{2} + 3y - 5 = x [/tex]

    [tex] \frac {d(y^{2}+3y-5)} {dx} = \frac {d(x)} {dx} [/tex]

    [tex] 2y \frac {dy} {dx} + 3 \frac {dy} {dx} = 1 [/tex]

    [tex] \frac {dy} {dx}(2y +3) = 1 [/tex]

    [tex] \frac {dy} {dx}(2y+3)-1 = 0 [/tex]

    [tex] (2y+3)-1 = \frac {dy} {dx} [/tex]

    I am not really sure if the last move is legal because 0 / anything is 0 so... is it more like this:

    [tex] \frac {dy} {dx}(2y +3) = 1 [/tex]

    [tex] \frac {dy} {dx} = \frac {1} {(2y+3)} [/tex]
     
    Last edited: Feb 24, 2010
  11. Feb 24, 2010 #10

    Dick

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    I think your main problem is just that your algebra is a little shakey. If it were more solid, you would know what's right and what's wrong. For the second one, this is correct,
    [tex]
    \frac {dy} {dx} = \frac {1} {(2y+3)}
    [/tex]
    for the first one this is correct,

    [tex]
    \frac {dy} {dx} = - \frac {1} {(x+y)^{2}} * (1 + \frac {dy} {dx})
    [/tex].

    Now you have to expand it and move all the dy/dx stuff to one side and everything else to the other side. The other stuff is just plain wrong algebraically, and as part of your algebra practice, you should tell me why they are wrong.

    For example, why is going from here,

    [tex]
    \frac {dy} {dx}(2y+3)-1 = 0
    [/tex]

    to here,

    [tex]
    (2y+3)-1 = \frac {dy} {dx}
    [/tex]

    just plain dead wrong.
     
    Last edited: Feb 24, 2010
  12. Feb 24, 2010 #11
    Well to answer the last question you posed, the best answer I can give you as to why its wrong is that 0 divided by anything is going to 0, not my dy/dx, maybe there is more that I forgot as to why it is not legal to make that move.

    Also after expanding the term in the first problem I gave you I get:

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

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

    but after this step if I divide to move the dx/dy on the right to the left it will leave me with 1 on the left which is not correct either....

    Also you are quite right about my algebra skills, algebra II is the last class I took in high school and then started teaching myself trig, but there was about a 4 year gap in between algebra II and starting trig... but thanks for all the help man!
     
  13. Feb 24, 2010 #12

    Dick

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    Ok, simpler question in the same form. Why don't x*a-1=0 and a-1=x have the same solution? That is essentially what you did with the question I posed. The simple answer to why it is illegal to make that move is that there is no legal way to make that rearrangement. Whatever is not legal is PROHIBITED. Besides, they give you different solutions for x. You have got to get those algebra skills up. Your math future depends on it.

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

    goes to,

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

    as a first step. Now can you solve for dy/dx?
     
    Last edited: Feb 24, 2010
  14. Feb 24, 2010 #13
    not exactly no... I briefly explored that option before posting my last posting, I have 'explored' it more since you confirm that was the direction I was suppose to go in but things get convoluted quickly here for me...

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

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

    then the large fraction on right becomes -1 and i think you could go:

    [tex] \frac {dy} {dx} (1+1) = -1 [/tex]

    and I could ultimately end up with dy/dx = -1/2 but I have no x or y variables left and again must be doing it wrong... sorry I am not catching on here man
     
  15. Feb 24, 2010 #14

    Dick

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    Ok. Your algebra, in fact, sucks. Big time. The left side becomes dy/dx*(1+1/(x+y)^2)). Just factor out the dy/dx. That is one of the few legal things you can do. That equals -1/(x+y)^2. Now you have an equation that is schematically a*(dy/dx)=b. Solve for dy/dx. And don't use voodoo.
     
    Last edited: Feb 24, 2010
  16. Feb 24, 2010 #15
    Alright man, I got to say, I don't know how I missed that, I'm usually pretty comfortable with factoring...but anyway I think this is right then....

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

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

    yay or nay?
     
  17. Feb 24, 2010 #16

    Dick

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    Yay. You can simplify that a little bit. If you dare...
     
  18. Feb 24, 2010 #17
    I'll give it a shot, maybe find some redemption, although it could go terribly wrong as well, haha

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

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

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

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

    that's what I got, I don't think there is any voodoo in there to speak of but as of right now, that confidence is kind of shaken
     
    Last edited: Feb 24, 2010
  19. Feb 25, 2010 #18

    Dick

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    Way wrong. Here's a trick that might help. Replace (x+y) with a number, like 2. Now figure out the value of the right side of each line. Where did it change from -1/5, the starting value to 3/4 the ending value? How many other values are in between? Where are the bad steps?
     
    Last edited: Feb 25, 2010
  20. Feb 25, 2010 #19
    ok how about this:

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

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

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

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

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

    I checked for consistency in the results all the way through, so I think that this is correct... NO VOODOO
     
    Last edited: Feb 25, 2010
  21. Feb 25, 2010 #20

    Dick

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    Now that's correct. No voodoo this time. If you put (x+y)=2, then the first line is -1/5 and the last line is -1/5. That's a pretty good check against voodoo, try checking numbers for a while, until you feel more confident.
     
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