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Leibnitz's notation and derivatives of implicit functions

  1. Aug 26, 2014 #1
    First off: I think I understand the chain rule and how it derives from
    [tex]
    \lim_{h \to 0} \frac{ f(x+h)-f(x)}{h}
    [/tex]
    and how to apply the chain rule when taking the derivative of an implicit function. The textbook I am reading Applied Calculus (by B. Rockett) uses the following example on differentiating implicitly:

    we are given the following function
    [tex]x^2+y^2 = 25;[/tex] We differentiate both sides with respect to [itex]x[/itex]:
    [tex]\frac{d}{dx}x^2 + \frac{d}{dx} y^2 = \frac{d}{dx} 25[/tex] then comes the part that confuses me. Taking the derivative of [itex]x^2[/itex] and [itex]25[/itex] is not the problem, the thing that I cant seem to get into my head is that middle part. OK [itex]y[/itex] is a function of [itex]x[/itex], so we ca apply the power rule (on the right).
    [tex]2x+ 2y \frac{dy}{dx} = 0;\hspace{20mm} \frac{d}{dx} y^n = n \cdot y^{n-1} \frac{dy}{dx}[/tex] 1. but how does the [itex]y[/itex] in dy/dx come about on the right side of the equation? why isn't it simply [itex] 2y \frac{d}{dx} [/itex] (without the y)?
    2. how can one express [itex]\frac{dy}{dx}[/itex] using limits as in the first equation in this post?
    3. what does 'with respect to _' mean formally?
     
    Last edited: Aug 26, 2014
  2. jcsd
  3. Aug 26, 2014 #2

    ShayanJ

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    I have two answers for you:

    1- The proof of chain rule.
    2- Chain rule is very intuitive. Imagine a drug company. It produces pills and packs them in two different boxes with different sizes and stores them. There is this storekeeper who should take care of the increases and decreases in the number of stored pills. He decides to only write changes in the number of pills and calculate the number of pills from those numbers at the end of the week. He also decides to write the changes only in terms of smaller boxes even if the pills are taken or brought in bigger boxes. So the change in the pills is equal to the product of number of boxes and the number of pills in a box(if the box is small). But what if the box is big? Then he should calculate the product of the number of big boxes and the number of pills in each small box and the number of small boxes in each big box . Got it?
     
  4. Aug 26, 2014 #3
    1. First of all ask yourself what would ##\frac{d}{dx}## mean. You can think of the derivative as an "action", an operator that must be applied to some function to get its meaning. A "derivative" without a function to derive is meaningless (at least in this kind of applications).
    To get that expression of the derivative of ##y^2## you can think of it as the composition of two functions: ##y^2(x) = g(y(x))## where ##g(x)=x^2##. Now if you apply the chain rule in the standard way what do you get?

    2. Exactly as in the limit that you wrote at the beginning of your post, writing ##y## instead of ##f##

    3. That depends on the context. From a mathematical point of view, it doesn't mean much in the case of functions of a single variable. If you have some ##f(x)## (function only of ##x##) then the only thing you can do is differentiate this function. You can of course say that you are "differentiating ##f## with respect to ##x##" but it is exactly the same thing as "differentiating ##f##".
    In the case of functions of many variables, you can have partial derivatives. Given a function of (say) two variables ##f(x,y)## you define the partial derivative with respect to ##x## as:
    [tex] \frac{∂f}{∂x}(x,y) \equiv \lim_{h \to 0} \frac{f(x+h,y)-f(x,y)}{h} [/tex]
    and similarly for the partial derivative with respect to ##y##.
    And this is (kind of) all from the mathematical standpoint. In the physical language the thing gets somewhat messier: given that the above said is still valid, we often talk about explicit and total derivatives with respect to some variable.
    For example if you have a function ##f(x,t)## representing the force acting on some point particle at the time ##t## when the particle is at the position ##x##, and if you know the trajectory of that point particle as a function ##x(t)## giving you the position of the particle at each instant of time, you have the explicit derivative of ##f## with respect to ##t##, which is the partial derivative of ##f## in the sense given above. But you can also talk of the total derivative of ##f## with respect to ##t##, which would be the derivative (note: i'm not saying partial derivative this time) of ##f## with respect to ##t##, when thinking of ##f## as a function of the single variable ##t##.
    More precisely, this means taking the derivative of the function ##t \mapsto f(x(t),t)##, i.e. of the function obtained by taking the composition of ##f## with ##x## expressed as a function of ##t##.
     
    Last edited: Aug 26, 2014
  5. Aug 26, 2014 #4

    Fredrik

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    What's confusing here is that the symbol y denotes both a real number and a function. Let's change the notation for the function to g. Since ##x^2+y^2=25## and ##y=g(x)##, we have ##x^2+g(x)^2=25##. Now you probably see that if you differentiate the left-hand side with respect to x, you get ##2x+2g(x)g'(x)##.

    If we now go back to using the symbol y both for g(x) and for g, this becomes ##2x + 2y \frac{dy}{dx}##.

    Also, it couldn't possibly turn out to include a factor of d/dx (with no function symbol to the right of it), because the derivative of a function that takes real numbers to real numbers, is a function that takes real numbers to real numbers, not a function that takes differentiable functions to functions.

    $$g'(x)=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$$ If we use the symbol y for both g(x) and g, this turns into
    $$y'(x)=\lim_{h\to 0}\frac{y(x+h)-y(x)}{h}.$$ We can write dy/dx instead of y'(x) without causing any confusion. We could write y instead of y(x), but it would probably be confusing to do so, and there's no good way to rewrite ##y(x+h)##.

    It's a way to specify what function you're supposed to take the derivative of. If you're asked to differentiate ##x^2\sin y## with respect to ##x## (and there's no constraint like ##x^2+y^2=25## that makes it possible to determine the value of one of the variables from the value of the other), you're supposed to find ##f'(x)##, where f is the function defined by ##f(s)=s^2\sin y## for all s, or equivalently, to find ##D_1g(x,y)##, where g is defined by ##g(s,t)=s^2\sin t## for all s,t. Similarly, if you're asked to differentiate ##x^2\sin y## with respect to ##y## (and there are no constraints), then you're asked to find ##h'(y)## where h is defined by ##h(s)=x^2\sin s## for all s.
     
    Last edited: Aug 26, 2014
  6. Aug 26, 2014 #5
    thank you for your posts, all the answers help, reading the answers written in different ways really gives more perspective on Leibniz's notation than a single answer would have. I still have a few questions anyhow, (it takes quite a while to formulate them so Ill just post them one at a time):

    Fredrik, if [itex]y=g(x)[/itex] then it follows from the given equation that [itex]\hspace{1mm}g(x)=x^2[/itex] and [itex]\hspace{1mm}y(x)=\sqrt{x^2+25}[/itex] this violates the two point rule however, but in any case the derivative of these composite functions is:[tex]

    \frac{d}{dx} g(y(x)) = g'(y(x)) \cdot y'(x) \hspace{5mm} = \hspace{5mm} 2 \cdot \sqrt{x^2+25} \cdot \frac{x}{ \sqrt{x^2+25} }\hspace{5mm} = \hspace{5mm} 2x

    [/tex]where is the mistake in my reasoning if I simply applied the chain rule?
     
  7. Aug 26, 2014 #6

    Fredrik

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    My g isn't the same function as the one that glance denoted by g. His g is such that ##g(t)=t^2## for all t. Mine is such that ##g(t)=\sqrt{25-t^2}## for all t.

    I'm not sure what you're doing there actually. What I can tell you for sure is that it doesn't follow from ##x^2+y^2=25## and ##y=g(x)## that ##g(x)=x^2##. In fact, what we get is that ##x^2+g(x)^2=25##. This implies that ##g(x)=\sqrt{25-x^2}##.
     
    Last edited: Aug 26, 2014
  8. Aug 26, 2014 #7

    Fredrik

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    You can start by solving ##x^2+y^2=25## for y, and then taking the derivative:
    $$x^2+y^2=25~~ \Rightarrow~~ y=\sqrt{25-x^2}~~ \Rightarrow~~ \frac{dy}{dx} =\frac{d}{dx}\sqrt{25-x^2} =\frac{1}{2\sqrt{25-x^2}}(-2x) =-\frac{x}{y}.$$ There's nothing wrong with this, but you're studying implicit differentiation, and the above is as explicit as it gets. The "implicit" option is to just differentiate both sides with respect to x, and then solve for dy/dx:
    $$x^2+y^2=25~~ \Rightarrow~~ 0=\frac{d}{dx}25 =\frac{d}{dx}\left(x^2+y^2\right) =2x+2y\frac{dy}{dx}=0~~ \Rightarrow~~ \frac{dy}{dx}=-\frac{x}{y}.$$
     
  9. Aug 26, 2014 #8
    Maybe you meant to write ##g(x)=\pm\sqrt{25-x^2}## and ##y=\pm\sqrt{25-x^2}## or ##|g(x)|=\sqrt{25-x^2}## and ##|y|=\sqrt{25-x^2}##?
     
  10. Aug 26, 2014 #9

    Fredrik

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    I was lazy and only bothered to write down equalities that are accurate on some part of that circle. I should have done what you did, or at least avoided implication arrows and used language that indicated that I was being sloppy.
     
    Last edited: Aug 26, 2014
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