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Math Help - Implicit Diff Problem. Whao made this stuff up

  1. #1
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    Implicit Diff Problem. Whao made this stuff up

    I have tried so many ways and I am sure I am over analyzing these problems but I dont know if I am suppose to use the Chain Rule Product Rule, I am becoming more confused and just need a step by step visual to get me jump started. If you have time could you please explain why each thing takes place. If not I understand, anyway here is the question.

    Find DxY by Implicit Diff

    xy + sin(xy) = 1
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  2. #2
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    Re: Implicit Diff Problem. Whao made this stuff up

    Quote Originally Posted by psilver1 View Post
    I have tried so many ways and I am sure I am over analyzing these problems but I dont know if I am suppose to use the Chain Rule Product Rule, I am becoming more confused and just need a step by step visual to get me jump started. If you have time could you please explain why each thing takes place. If not I understand, anyway here is the question.

    Find DxY by Implicit Diff

    xy + sin(xy) = 1
    \frac{d}{dx}(xy)+\frac{d}{dx}\sin(xy)=\frac{d}{dx}  1

    y \frac{d}{dx}x+x\frac{d}{dx}y+\cos(xy)\frac{d}{dx}(  xy)=0

    y \frac{d}{dx}x+x\frac{d}{dx}y+\cos(xy)(y\frac{d}{dx  }x+x\frac{d}{dx}y)=0

    y+x\frac{dy}{dx}+\cos(xy)(y+x\frac{dy}{dx})=0

    Now just solve for the derivative.
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    Re: Implicit Diff Problem. Whao made this stuff up

    For the first term, we may use the product rule, for the second term the chain rule, then the product rule, and the right side is of course zero:

    x\cdot\frac{dy}{dx}+y+\cos(xy)\left(x\cdot\frac{dy  }{dx}+y \right)=0

    \left(x\cdot\frac{dy}{dx}+y \right)(1+\cos(xy))=0

    What does this imply?
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    Re: Implicit Diff Problem. Whao made this stuff up

    So (xy' + y)(xy'cos(xy) + ycos(xy))=0
    sorta stuck does the xy' factor out. (xy' + y)(cos(xy) + ycos(xy))=0

    thanks for the help so far. the above is where I am screwing this thing up. Mark I dont see where the +cos(xy)[x * dy/dx + y] turns into 1 + cos(xy). Btw I am sure you are right, i just dont understand where it comes from is what I am trying to say
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    Re: Implicit Diff Problem. Whao made this stuff up

    Quote Originally Posted by psilver1 View Post
    So (xy' + y)(xy'cos(xy) + ycos(xy))=0
    sorta stuck does the xy' factor out. (xy' + y)(cos(xy) + ycos(xy))=0

    thanks for the help so far. the above is where I am screwing this thing up. Mark I dont see where the +cos(xy)[x * dy/dx + y] turns into 1 + cos(xy). Btw I am sure you are right, i just dont understand where it comes from is what I am trying to say
    Look closely, Mark simply factored the whole thing. It's correct.
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    Re: Implicit Diff Problem. Whao made this stuff up

    Ahh I see it. So is 1+cos(xy) and identity. Sorry for the questions just what to clear my mind.
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  7. #7
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Implicit Diff Problem. Whao made this stuff up

    xy+sin(xy)-1 as surface:


    Attached Thumbnails Attached Thumbnails Implicit Diff Problem.  Whao made this stuff up-implicit-differentiation.png  
    Last edited by MaxJasper; October 4th 2012 at 09:24 PM.
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Implicit Diff Problem. Whao made this stuff up

    It is simply a factor we may ignore, as it doesn't involve \frac{dy}{dx}, and we know it cannot be zero, since this would imply:

    xy=(2k+1)\pi

    But, if this is true, then the original equation becomes:

    xy=1

    Which is a contradiction.
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    Re: Implicit Diff Problem. Whao made this stuff up

    So ignoreing the (1+ cos(xy). We are left with xy' + y = 0 which would give y'= - y/x. Which is the correct answer but where does the cosxy move away too.
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Implicit Diff Problem. Whao made this stuff up

    It doesn't move away, we just ignore it as a factor that cannot be zero.

    If you use the method suggested by TheEmptySet, you will find this factor is divided out or cancels. I highly suggest you try this method as well.
    Last edited by MarkFL; October 4th 2012 at 10:04 PM.
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