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Thread: Another Implicit differenriarion

  1. #1
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    Another Implicit differenriarion

    Lets see if I learned from the last one.

    2x^2y=sin2x

    \frac{d}{dx}[2x^2y] = \frac{d}{dx}[sin2x]

    4xy\frac{dy}{dx}[2x^2]=\frac{dy}{dx}[cos2x]\frac{dy}{dx}[sin2]

    \frac{dy}{dx}(2x^2-cos2x-sin2) = -4xy

    \frac{dy}{dx} = -\frac{4xy}{2x^2-cos2x-sin2}
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  2. #2
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    Should be product rule again on left-hand side, and I'm not at all sure what you're aiming at there on the right (should be chain rule). Balloons approaching...

    Edit:

    I see now that you were trying for the product rule but forgot the plus sign. Anyway...



    Just in case it helps to see the formulae as well...



    and




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    Last edited by tom@ballooncalculus; Aug 25th 2009 at 02:08 PM. Reason: see now
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  3. #3
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    2x^2y=sin2x

    \frac{d}{dx}[2x^2y] = \frac{d}{dx}[sin2x]

    4xy+\frac{dy}{dx}[2x^2]=\frac{dy}{dx}[2cos2x]

    \frac{dy}{dx}(2x^2-2cos2x) = 4xy

    \frac{dy}{dx} = \frac{4xy}{2x^2-2cos2x}
    Last edited by drkidd22; Aug 25th 2009 at 02:24 PM.
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  4. #4
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    You still want to see a dy/dx on the right hand side? Not so...



    LHS good.


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  5. #5
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    I keep it on both sides so that I later remember what it was, then I move everything to one side.
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  6. #6
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    Yes but don't put it in if it doesn't belong... are you confusing dy/dx with d/dx? It's leading you down a wrong turn - your next line is way off

    Edit:

    Your factorising and so on are fine - subject the bottom row of my pic to the same processes and you should arrive at the correct derivative. Then have a think about why you were taking a different turn... probably the d/dx notation, which actually just means 'derivative (with respect to x) of...' (You don't need it anyway if you draw the picture!)
    Last edited by tom@ballooncalculus; Aug 25th 2009 at 02:50 PM.
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  7. #7
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    yeah, I think I'm confusing it with d/dx. Which line is way off? I though I had it right already
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  8. #8
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    I get confused when doing this derrivative.

    \frac{d}{dx}[sin2x]

    explain please, I think you already did, but my brain is not absorbing it.
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  9. #9
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    \frac{d}{dx}\sin(2x) just means the derivative of sin(2x).

    The chain rule says that the derivative of sin(anything) is cos(the same thing) times the derivative of the thing...



    Try to read d/dx as just saying 'the derivative of' - or use a picture and do without it, because d/dx of anything is whatever's in the balloon below...
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  10. #10
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    <br />
4xy+\frac{dy}{dx}[2x^2]=2cos2x

    \frac{dy}{dx} = \frac{2cos2x-4xy}{2x^2}<br />
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  11. #11
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    Yep
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  12. #12
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    I know why I got confused when doing that part. It now shouldn't happen again.

    Thanks alot
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