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Partial derivative

  1. Dec 6, 2015 #1
    x2y2 + (y+1)e-x=2 + x

    Defines y as a differentiable function of x at point (x, y) = (0,1)

    Find y′:

    My attempt:

    ∂y/∂x =2xy3 + (-y-1)e-x=1

    ∂y/∂y = 3x2y2 - e-x=0

    Plugging in for x and y ⇒

    ∂y/∂x = -3

    ∂y/∂x = -1

    For some reason I think y′ is defined as
    (∂y/∂x) /(∂y/∂y) = 3

    At leas this give the right answer (from solutions), but I am uncertain so I hope someone can correct me where I am wrong.
     
  2. jcsd
  3. Dec 6, 2015 #2

    Mark44

    Staff: Mentor

    In future posts, please do not delete the homework template.

    No, not even close.
    First, you shouldn't be using partial derivatives, as the problem statement says that y is a differentiable function of x.
    Second, and more important, to find dy/dx, you need to use implicit differentiation.
     
  4. Dec 6, 2015 #3
    I looked up implicit differentiation in my book ( thanks for pointing me in the right direction), but it seems like that is what I have been doing even without knowing it. What I did was correct and the answer is correct.
     
  5. Dec 6, 2015 #4

    Mark44

    Staff: Mentor

    The answer might possibly be correct, but your work is not correct.

    Starting with this equation:
    ##x^2y^2 + (y + 1)e^{-x} = 2 + x##
    and differentiating implicitly with respect to x, we get
    ##2xy^2 + 2x^2y*y' - (y + 1)e^{-x} + y' e^{-x} = 1##
    To continue, put all of the terms involving y' on one side, and all the others on the other side, and solve algebraically for y'.
     
  6. Dec 6, 2015 #5
    The formula I found in the book says:

    Slope of a level curve ( which is what I have since 2 is a constant)

    F(x,y) = c ⇒ y′ = F`1 (x, y) / F`2 (x, y)
    F`2 (x, y) ≠ 0

    which is exactly what I did. Maybe you would get the same answer doing it your way, but what I did is correct as well.
     
  7. Dec 6, 2015 #6

    Mark44

    Staff: Mentor

    Here's what you posted earlier:
    The function for your level curve would be F(x, y) = C, or x2y2 + (y+1)e-x - x =2
    So the left side of the last equation if F(x, y), and C = 2
    2xy3 can't be right.
    Also, the above should be ∂F/∂x and below should be ∂F/∂y, since ∂y/∂y is 1.
    I think you stumbled onto the correct answer by accident, which is not a good way to do things. You lucked out that the point in question is (0, 1), so lots of terms drop out.
     
  8. Dec 6, 2015 #7
    You are right that it should be ∂F/∂x and below should be ∂F/∂y, since ∂y/∂y is 1 and also I wrote y3 instead of y2 which could be confusing. But as you said it doesent matter cus it drops out.
     
  9. Dec 6, 2015 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I hope you realize that it matters a lot. You would really see the difference if you started at a different point, such as ##(x,y) = (1,y_1)##, where
    [tex] y_1 = \frac{e^{-1}}{2} \left( -1 + \sqrt{12e^2 - 4e +1} \right). [/tex]
    Another point is that you might be marked wrong even if your answer is correct but your argument and formulas are incorrect.
     
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