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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    Find the slope of the tangent line to the curve

    points (1,4)


    i can get it to the point where you need to isolate the derivative, but im not sure how its done in this example
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  2. #2
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    {\color{blue}\sqrt{3x+2y}} + {\color{red}\sqrt{3xy}} = 6.75
    {\color{blue}(3x+2y)^{\frac{1}{2}}} + {\color{red}(3xy)^{\frac{1}{2}}} = 6.75

    Differentiating:

    \underbrace{{\color{blue} \frac{1}{2}(3x+2y)^{-\frac{1}{2}} \cdot (3 + 2y')}}_{\mbox{Chain rule}} + {\color{red} \frac{1}{2}(3xy)^{-\frac{1}{2}} \cdot } \underbrace{{\color{red}3 (y + xy')}}_{\mbox{Product Rule}} = 0

    Now you have to do a bit of algebra to isolate y' but it should be doable. Collect all the terms with y' on one side and everything else on the other. Factor out y' and you should be good. Then plug in (1,4) to your end result.
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  3. #3
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    thats exactly the point im stuck at. How do you isolate y' when both of them are in products?
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  4. #4
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    To make it clearer, let:
    a = {\color{blue} \frac{1}{2}(3x + 2y)^{-\frac{1}{2}} }
    b = {\color{red} \frac{1}{2}(3xy)^{-\frac{1}{2}} }

    So we have:
    a(3 + 2y') + 3b(y + xy') = 0
    3a + {\color{red}2ay'} + 3by + {\color{red}3bxy'}

    Notice how you can isolate y' now? You probably got confused underneath all that garbage going on eh?
    Last edited by o_O; April 12th 2008 at 02:21 PM. Reason: Too many typos : (
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  5. #5
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    I dont understand why a(3+2y') is 3a +2y' and not 3a +2ay'. The same for the latter example. Why is that so? You can just pull the y' out even if its part of a product?

    shouldnt b= 1/2(3xy)^-1/2
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  6. #6
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    Oh dear. I made so many typos.

    Yes, I did mean a(3x + 2y) = 3ax + 2ay'.

    Yes, I did mean b= 1/2(3xy)^-1/2.

    Sorry about that. I hope you get the idea though.
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  7. #7
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    can you go further and solve it? Ive tried numerous ways and ive asked multiple people and no1 can get the answer
    Last edited by EricK; April 12th 2008 at 02:52 PM. Reason: typo
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  8. #8
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    Keeping the same equivalences for a and b (kind of lazy):
    a(3 + 2y') + 3b(y + xy') = 0
    3a + 2ay' + 3by + 3bxy' = 0
    2ay' + 3bxy' = -3a - 3by
    y'(2a + 3bx) = -3(a + by)
    y' = \frac{-3(a + by)}{2 a+ 3bx}

    Then plug in x = 1, y = 4.
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  9. #9
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    once again though what happened to 2ay' and 3bxy'?
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  10. #10
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    Ah sorry, guess reading through the latex causes me to miss little things. anyway, I edited my post for ya
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  11. #11
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    there has to be something wrong with that unless im doing some little arithmatic error wrong. I'm doing this on a webwork system, so it tells me when its wrong.
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  12. #12
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    Most likely. It's pretty messy to plug in. What did you get for your answer?
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  13. #13
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    I've had so many answers... lets just say for arguements sake .4017475548, which is no way the answer.
    Ive had answers like that, negative answers, answers like 6.2, 5.45, etc
    Last edited by EricK; April 12th 2008 at 08:00 PM. Reason: soooooom many typos
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  14. #14
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    Well it should be negative. Ok well, I wanted to avoid typing it all out but ...

    I started from scratch and found y' and even did wrote it all neatly on paper to make sure I didn't make a mistake (which I hopefully didn't):
    y' = \frac{-3\left[(3x + 2y)^{-\frac{1}{2}} + y(3xy)^{-\frac{1}{2}}\right]}{2\left[(3x + 2y)^{-\frac{1}{2}} + \frac{3x}{2}(3xy)^{-\frac{1}{2}} \right]}

    Now, plugging in (1,4):

    y' = \frac{-3 \left[(3(1) + 2(4))^{-\frac{1}{2}} + (4)\left[3(1)(4)\right]^{-\frac{1}{2}} \right]}{2 \left[(3(1) + 2(4))^{-\frac{1}{2}} + \frac{3(1)}{2}(3(1)(4))^{-\frac{1}{2}} \right]}

    y' = \frac{-3\left[(11)^{-\frac{1}{2}} + 4(12)^{-\frac{1}{2}} \right]} {2 \left[ (11)^{-\frac{1}{2}} + \frac{3}{2}(12)^{-\frac{1}{2}}\right]}
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  15. #15
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    well i plugged it into my calc and i got -4.2166.....
    answer was wrong though
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