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Evaluating a derivative by partial differentiation proof

  1. Jun 4, 2015 #1
    1. The problem statement, all variables and given/known data

    Suppose we have an equation,
    ex + xy + x2 = 5
    Find dy/dx
    2. Relevant equations
    Now I know all the linear differentiation stuff like product rule, chain rule etc.
    Also I know partial differentiation is differentiating one variable and keeping other one constant.
    3. The attempt at a solution
    By my method I can evaluate like,
    Differentiating both sides
    ex + y + xdy/dx + 2x = 0
    and then rearrange to get dy/dx.

    But my question is
    When f(x,y) = 0
    How dy/dx = -δx/δy ?
    This is really a fast method as we don't have to rearrange terms.
     
  2. jcsd
  3. Jun 5, 2015 #2

    Svein

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    How do you explain that?

    I know that in some cases you have [itex]\frac{dy}{dx}=(\frac{dx}{dy})^{-1} [/itex], but it presupposes that y(x) is injective and that x(y) is meaningful and differentiable.
     
  4. Jun 5, 2015 #3
    Okay the function is continuous and differentiable everywhere.
     
  5. Jun 5, 2015 #4
    And sorry, it should be
    $$ dy/dx = \frac{-δf(x,y)/δx}{δf(x,y)/δy} $$
     
  6. Jun 5, 2015 #5

    SteamKing

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  7. Jun 5, 2015 #6

    Svein

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    That's better. And even correct since you start out with f(x, y) = 0. Then [itex] 0 = df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/itex].
     
  8. Jun 5, 2015 #7
    Because we do not have to rearrange terms in this partial derivative method and it is fast.
    Not getting this step,
    If f(x,y) = 0 then how df= 0.
    Note I have not taken classes of partial differentiation.
    I Only know this formula and applying because it is fast as compared to implicit differentiation.
    Can you give me some insight on deriving the formula?
     
  9. Jun 5, 2015 #8

    SteamKing

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    Dudn't seem to be faster, otherwise there wouldn't be this extended thread on getting all the terms correct.

    As far as not having to rearrange terms, are you sure you haven't "cherry-picked" your example function from the OP?
     
  10. Jun 5, 2015 #9

    Svein

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    If f = constant, then df = 0.
     
  11. Jun 5, 2015 #10

    Mark44

    Staff: Mentor

    Implicit differentiation is definitely the way to go here, not partial derivatives, as SteamKing suggests. You are way overcomplicating things by invoking partial derivatives
     
  12. Jun 5, 2015 #11
    For this question, it's easy to express y in terms of x and differentiate. You don't even need implicit differentiation.
     
  13. Jun 6, 2015 #12
    But I don't see it complicated if I know the formula.
    For example taking my example in OP
    ex + xy + x2-5 = 0
    Now, δf(x,y)/δx = ex + y + 2x
    δf(x,y)/δy = x
    So by formula dy/dx = -(ex + y + 2x)/x
    ---------

    f(x, y) = 0. Then [itex] 0 = df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/itex]
    I'm not understanding the red part. How we are arriving at that?
     
  14. Jun 6, 2015 #13

    WWGD

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    This is the definition of the differential of a function of two variables.

    http://en.wikipedia.org/wiki/Differential_of_a_function

    It is not an arbitrary definition; the differential is the best local-linear approximation to the change of
    a (differentiable)function, in a precise ## \delta - \epsilon ## sense.
     
  15. Jun 6, 2015 #14

    Mark44

    Staff: Mentor

    I think you are not understanding the formula you posted. We don't "arrive" at what you have in red, above. As already stated, that's the definition of the differential of f, with f being a function of two variables.

    From this definition, you can derive the formula you're using.
     
  16. Jun 6, 2015 #15
    Yes I am not understanding the formula.
    As I told you I have not taken a class of partial differentials yet.
    I mostly deal with dy, dx not δy and δx.
    Is ∂x and δx different?
     
  17. Jun 6, 2015 #16
    Now, I am not understanding the definition formula.
    What to do if you are not understanding definition and what to do if you know only the ordinary differentiation?
     
  18. Jun 6, 2015 #17
    If you only know ordinary differentiation, you had better not used this method. It's easy to get things wrong this way.

    For this question, it's really easy to express y in terms of x and differentiate.
     
  19. Jun 6, 2015 #18
    But I want to learn a new thing.
    The attempt I have shown in OP for implicit differentiation.
    Edit - moderator you can delete this and subsequent message , don't know why the messages repeated at same time.
     
  20. Jun 6, 2015 #19
    But I want to learn a new thing.
    The attempt I have shown in OP for implicit differentiation.
     
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