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Homework Help: Theoretic doubt about the definition of derivatives.

  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Hi, this is a question that has been bothering me for a while. (Im in calculus II at the moment)

    Why do i need to derivate some functions by definition and other times i dont? for example if somebody asks me to calculate the partial derivatives of a branch function in a a certain point, lets say (0,0).

    i know how to solve the exercices but i dont understand the concept.


    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 10, 2016 #2


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    In this generality, I would firstly ask, what are functions and why to consider them at all?

    In the end, they are (often) used to describe a process, either in time or the dependencies between different locations. Or simply the size of a physical property. Many functions are quite complicated and one would like to have an approximation which is easier to compute. But almost always it is about to find out the behavior, i.e. the amount of changes: is it defined, is it polynomial, exponential or constant? For both the derivative can be helpful, because it is a local linear approximation and therewith easy to compute, and it shows the amount of change in a certain direction. You might want to know something about the time, a pandemic spreads, or about the slope of your track you're going to ski downhill.
  4. Oct 10, 2016 #3
    hum... that seems kind of vague, i just wanted to know why in some cases i cannot calculate the partial derivates by the the rules and i need to use the definition: let me ilustrate with an example.... (Lets say i want to calculate the partial derivatives of f(x,y) and f(x,y) is:

    | (y³) / (x⁴ + y²) if( x,y) is not 0
    | 0 if (x,y) is 0
  5. Oct 10, 2016 #4


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    Well, it is always the usage of the definition. The rules are proven from the definition and provide a toolbox, which might apply to a problem or not.
    Asking why a certain wrench fits and others don't is a bit strange, isn't it?
  6. Oct 10, 2016 #5
    Rules are way faster than the definition,(if i want to be more efficient into solving the exercices i need to use them) but sometimes, as the case i posted i cannot use them. I just wanted to know why, im not always going to use the same wrench (definition) cause in the end i would lose time.
  7. Oct 10, 2016 #6


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    Sometimes the definition helps for special cases. For example, how would you prove that the derivative of ##f: x \mapsto |x|## does not exist in ## x = 0## using your rules?
  8. Oct 10, 2016 #7
    ok i think im starting to get it.
  9. Oct 10, 2016 #8


    Staff: Mentor

    Using the definition of the derivative is more like making a wrench each time you need to tighten a nut and bolt, and the various rules are like wrenches that have already been made.

    @Muradean, the word "derivate" is a word, but not one that is used in mathematics. To find the derivative of a function, you differentiate it. You don't "derivate" it.
  10. Oct 10, 2016 #9

    Ray Vickson

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    For any ##(x,y) \neq (0,0)## the function "behaves well", and so rules apply readily. Of course, the rules are derived from the definition, but are easier and faster to use than the definition itself.

    However, at certain points the "rules" can become problematic; expect trouble whenever you try to divide by 0, or whenever you reach the boundary of the domain of a root-function for a root < 1; for example, trouble can occur at ##x = 0## if your function contains ##\sqrt{x}## or ##x^{3/4}##, or ##x^p## for positive ##p < 1##. In your example ##f(x,y)##, we can expect trouble when ##x = y = 0##, because we would then be dividing by 0 if we tried to use the formula in line 1. Of course, when ##(x,y) = (0,0)## we use line 2, giving ##f(0,0) = 1##, so in the actual ##f## we are not dividing by 0 because we switch to a different formula.

    At a "troublesome" point, you need to fall back on the actual definition of the partial derivatives. By definition:
    $$f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{0-0}{h} = 0,$$
    $$f_y(0,0) = \lim_{h \to 0} \frac{f(0,h) - f(0,0)}{h} = \lim_{h \to 0}\frac{(h^3/h^2) - 0}{h} = \lim_{h \to 0} 1 = 1$$
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