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Math Help - Derivative of Implicit function, EXAM TOMORROW MORNING, PLEASE HELP!!!

  1. #1
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    Derivative of Implicit function, EXAM TOMORROW MORNING, PLEASE HELP!!!

    -3x^2+1xy-1y^3=-15

    I have to find the slope of the tangent line at the point of (-2,1)

    Now from my work, the derivative of this function is

    dy/dx = (-6x+y)/(3+3y^2)

    Therefore the slope = 13/6

    But the answer ain't right, why?

    If anyone can help before I head off to the exam tomorrow morning at 9:30am I will really really really really appreicate it!!!

    THANK YOU VERY MUCH!!!
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  2. #2
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    Quote Originally Posted by tttcomrader
    -3x^2+1xy-1y^3=-15

    I have to find the slope of the tangent line at the point of (-2,1)

    Now from my work, the derivative of this function is

    dy/dx = (-6x+y)/(3+3y^2)

    Therefore the slope = 13/6

    But the answer ain't right, why?

    If anyone can help before I head off to the exam tomorrow morning at 9:30am I will really really really really appreicate it!!!

    THANK YOU VERY MUCH!!!
    Hello,

    to derivate an implicite function use:
    \frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}

    With your problem:
    \frac{\partial f}{\partial x}=-6x+y
    \frac{\partial f}{\partial y}=-3y^2+x

    Greetings and good luck with your exam!

    EB
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  3. #3
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    earboth, he said implicit differentiation, not partial differentiation. hence it doesn't say
    f(x,y) = -3x^2 + xy - y^3 + 15 or anything like that

    anyway ttt, your answer shows you've got some idea how to go about things but its not quite right. we have
    -3x^2 + xy - y^3 = -15
    differentiating both sides... i will write y' instead of \frac{dy}{dx} to save time.
    i will take you through term by term
    for -3x^2, that differentiates to -6x, which i'm sure you already know.
    now next i will look at y^3
    if we had
    f(x) = y^3
    using the substitution h=y
    f(x) = h^3
    then
    \frac{df}{dh} = 3h^2
    also from the h=y we have
    \frac{dh}{dx} = \frac{dy}{dx}
    so \frac{df}{dh} \times \frac{dh}{dx} = 3y^2 \times \frac{dy}{dx}
    notice the dh's cancel
    so differentiating y^3 implicitly with respect to x gives  3y^2 y'
    which you might have also known.
    where to do it quickly you are basically differentiating it with respect to y to get 3y^2. then multiplying it by \frac{dy}{dx} to give 3y^2y'

    now for the xy part you have to use the product rule
    let
    g = x
    and
    h=y
    so
    g' = 1
    h' = y'
    then \frac{d(xy)}{dx} = gh' + hg' = xy' +y
    so to do it quickly when doing something with an x term times a y term you first differentiate with respect to x then multiply it by the y term... so to give you 1 \times y then you add that to differentiating with respect to y implicitly multiplying by the x term which gives x \times \frac{dy}{dx} or xy'
    so in the original problem:
    -3x^2 + xy - y^3 = -15
    this becomes
    -6x + y + xy' - 3y^2y' = 0
    (the 15 differentiates to become 0 which you know)
    so rearranging this gives
    xy' - 3x^2y' = 6x - y
    then
    y' = \frac{6x-y}{x-3y^2}
    then you can put in x=-2, y=1
    y' = 13/5

    by the way i apologize with my poor attempts at explaining whats going on in these examples. i just wanted to try and explain rather than merely give answers. i would appreciate it if somebody who knows things better would come along and clear them up for me.
    Last edited by Aradesh; April 4th 2006 at 04:30 AM.
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  4. #4
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    Thank you, that is really helpful, I think I'm ready for the exam now.

    I"m heading out for it, let's hope I will come back smiling.

    Again, thanks for all of your helps!

    *Update*

    I just got back from school, the exam is much easier than the homework, I'm pretty confidence I made an A on that.

    Thank you!
    Last edited by tttcomrader; April 4th 2006 at 02:11 PM.
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