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Math Help - Implicit differentiation

  1. #1
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    Implicit differentiation

    Q: Use implicit differentiation to find an equation of the line tangent to the curve yx^2 + xy^2 = 6 at the point (1, 2).

    Here's my answer... it is right?

    2x(d/dx) + 2y(d/dx) = 6(d/dx) = 0 --> (d/dx)(2x + 2y) = 0 --> dy/dx = 0/(2x + 2y) = 0 = m

    y = mx + b = 0x + b = b = 2 --> y =2
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  2. #2
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    Just in case a picture helps...



    Spoiler:


    Then solve the bottom row for dy/dx and plug in the x and y values for the given point to evaluate m at the point.


    ... is the product rule, and...



    ... the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    _________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by satx View Post
    Q: Use implicit differentiation to find an equation of the line tangent to the curve yx^2 + xy^2 = 6 at the point (1, 2).

    Here's my answer... it is right?

    2x(d/dx) + 2y(d/dx) = 6(d/dx) = 0 --> (d/dx)(2x + 2y) = 0 --> dy/dx = 0/(2x + 2y) = 0 = m

    y = mx + b = 0x + b = b = 2 --> y =2
    note that you need the product rule here! y is treated as a function of x. you are to differentiate as usual, but when you differentiate a y-term, you must attach dy/dx to it (this follows from the chain rule). Hence,

    yx^2 + xy^2 = 6

    \Rightarrow 2xy + x^2 \cdot \frac {dy}{dx} + y^2 + 2xy \cdot \frac {dy}{dx} = 0

    Now solve for \frac {dy}{dx} and plug in x = 1,~y = 2. This will give you the m (that is, the slope) for your line

    Then use y - y_1 = m(x - x_1) with (x_1,y_1)= (1,2) and the m found above. Solve for y and that's the equation of the tangent line
    Last edited by CaptainBlack; April 5th 2010 at 01:17 AM. Reason: removed comment on post I have deleted to avoid confusion
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